Symmetric latin square and complete graph analogues of the evans conjecture

Symmetric Latin Square and Complete Graph Analogues of the Evans Conjecture

Lars Dsvling Andersen Department of Mathematics and Computer Science, Institute of Electronic Systems, Aalborg University, 9220 Aalborg 0, Denmark

A. J. W. Hilton Department of Mathematics, West Virginia University, Morgantown, WV 26506 and Department of Mathematics, University of Reading, P. 0. Box 220, Reading RG6 2AX, England

ABSTRACT

With the proof of the Evans conjecture, it was established that any partial latin square of side n with a most n - 1 nonempty cells can be completed to a latin square of side n. In this article we prove an analogous result for symmetric latin squares: a partial symmetric latin square of side n with an admissible diagonal and at most n - 1 nonempty cells can be completed to a symmetric latin square of side n. We also characterize those partial symmetric latin squares of side n with exactly n or n + 1 nonempty cells which cannot be completed. From these results we deduce theorems about completing edge-colorings of complete graphs Kzm and Kzm - 1 with 2m - 1 colors, with m + 1 or fewer edges getting prescribed colors. 0 1994 John Wiley & Sons, Inc.

1. INTRODUCTIQN

The Evans conjecture of 1960 [12] for latin squares was proved by Smetaniuk in 1981 [21] and in a strengthened form, independently and by a completely different method, also by the authors [5]. Later, the first author proved a further strengthening [ 2 ] . Here we prove a similar result for symmetric latin squares.

Apartial latin square of side n on the symbols s1,. . . , s,, is an n X n matrix in which each cell is either empty or contains one of the symbols $1,. . . , s,, and, furthermore, no symbol occurs twice in any row or twice in any column. It is a latin square if there are

0 1994 John Wiley & Sons, Inc. CCC 1063-X539/94/040197-56

h m m l of Combinatorial Designs, Vol. 2, NO. 4 (1994) 197

198 ANDERSEN AND HILTON

no empty cells, The conjecture of Evans was that any partial latin square of side n with at most n - 1 nonempty cells can be completed to a latin square of side n . In [ 5 ] , we strengthened this and proved the following theorem.

Theorem 1. For all n 2 1, a partial latin square of side n with at most n nonempty cells can be completed to a latin square of side n if and only if it is not of the form of any of Types 1, 2 or 3 of Figure 1.

No doubt the meaning of the phrase “of the form of’ is self-evident, but, more formally, a partial latin square is of the form of one of these Types if it can be transformed to a latin square of Figure 1 (for some x) by permuting rows, permuting columns, and relabeling symbols.

The strengthening obtained by the first author was the following.

Theorem 2. For all n 2 1, a partial latin square of side n with at most n + 1 nonempty cells can be completed to a latin square of side n if und only if it tieither has n of its nonempty cells forming a square of Type 1,2 or 3, nor is of any of Types 4- 13 of Figure 2.

The idea of formulating and proving a similar result for symmetric latin squares has occurred to various people. This work began when it was suggested by Lindner at the workshop “Latin squares: Their construction and application” at Simon Fraser University in Vancouver in 1983. We quickly obtained such a result for the analogous problem for edge-coloring complete graphs (see Theorem 9), but we have since spent a considerable amount of time extending our result to symmetric latin squares, trying to shorten the very long proof, and, at the same time, strengthening our results. In the theorem we prove here, we assume n + 1 preassignments-the analogue of Theorem 2.

Before we state the result we recall that in a symmetric latin square of even side, each symbol occurs an even number of times on the diagonal, and in a symmetric latin square of odd side, each symbol occurs exactly once on the diagonal. Apartial symmetric latin square is, of course, a partial latin square which is also symmetric. We shall call the diagonal of a partial symmetric latin square of side n admissible if the number of symbols occurring on the diagonal a number of times not congruent to n modulo 2 is less than or equal to the number of empty diagonal cells (so that the parity can be made right for every symbol with the “wrong” parity). Let t(i) denote the number of times that a symbol i occurs on the diagonal. Then, if the symbol set is (1,. . . , n}, the condition

1 * * * x

1$ Type 1 (1s x < n)

Type 2 ( lsx < n)

Fig. 1. The Noncompletable Partial Latin Squares of Side n with n Nonempty Cells.

EVANS CONJECTURE 199

Type 4 (n 2 3) It n-2 n-1

Type 7 (n 2 4)

- Type 10 (n 2 5)

Type 5 (n 2 3)

[ n-3

Type 8 (n 2 4)

Type 11 (n 2 5)

I y *=.

I m 1 Type 6 (n Z 3)

Type 9 (n 2 4)

2 3 m 4 5

Type 12 (n 2 5)

Type 13 (n = 4)

Fig. 2. Noncompletable Partial Latin Squares of Side n with n + 1 Nonempty Cells.

for admissibility can be written

I{i I t(i) p n (mod 2)}1 5 n - t ( i ) . i=l

For odd n, this means that a diagonal is admissible if and only if no symbol occurs more than once on it. Obviously, a partial symmetric la th square cannot be completed if its diagonal is not admissible.

Our main result is the following.


Theorem 3. Let n 2 1 and let P be a partial symmetric latin square of side n with admissible diagonal. Let c be the number of nonempty cells of P.

I f c 5 n - 1, then P can be completed to a symmetric latin square of side n. If c = n, then P can be completed to a symmetric latin square of side n if and only if

P is not of the form of any of the Types E l , 0 1 or 0 2 of Figure 3. If c = n + 1, then P can be completed to a symmetric latin square of side n if and

only if P is neither of the form of any of the Types E l , 0 1 or 0 2 with a further diagonal cell filled, nor ofthe form of any of the Types E2, E3, 0 3 , 5A or 5B of Figure 3.

even - even even

... _I ~-

Type El (n even) c = n

even - Type E2 (n even) c = n + i

Type 01 (n odd) c = n

Type 5A (n = 5) c = n + l

1

Type 02 (n odd) c = n

Type 5B (n = 5 ) c = n + l

even

Type E3 (n even) c = n + l

"1 Type 03 (n odd)

c = n + l

Fig. 3. Cells.

Noncompletabie Partial Symmetric Latin Squares of Side n with at most n + 1 Nonempty


When we use the phrase “ P is in the form of Q” in Theorem 3, we mean that P can be transformed to Q by carrying out the same permutation on the rows and the columns, and by renaming the symbols (in a one-to-one manner, naturally).

It is easy to see that the partial latin squares in Figure 3 cannot be completed symmetrically. In Types El and 01 the symbol 1 needs an extra diagonal occurrence but cannot get it; in Types E2 and 0 2 it is impossible to make the symbol 1 occur in the first row; in Types E3 and 03 the symbol 1 must occur in the last diagonal cell, but the square obtained in this way contains one of Type E2 or 02, respectively; finally we leave it to the reader to check that Types 5A and 5B cannot be completed.

We prove Theorem 3 in Section 7. An important special case was, however, proved in an earlier article [6]. This case, where there is at least one nonempty cell in each row and column of the partial symmetric latin square, is not open to the general method of the present article.

Theorem 4 [6]. one occupied cell in each row and in each column.

Theorem 3 holds for any partial symmetric latin square P with at least

There are other results, also some due to the authors, that we shall need in the proof; we state them in the context where they are discussed. In Section 2 we present a graph theoretic reformulation of Theorem 3, as well as some consequences.

Finally we remark that the basic method of proof in this article is the same as in our proof of the Evans conjecture in [5 ] . With partial latin squares, the roles of rows, columns, and symbols can be interchanged when convenient. Much of this interchangeability is lost with symmetric latin squares, and this accounts in good measure for the increased length of this article. We mentioned earlier that the elegant proof by Smetaniuk 1211 of the Evans conjecture, later strengthened by Damerell [ll], is entirely different from ours. It is our strong impression that it cannot be adapted to give a proof of the results in this article.

2. COMPLETING PARTIAL EDGE-COLORINGS OF COMPLETE GRAPHS

In this section we discuss edge-colorings of graphs. There are two important reasons for this. Firstly our proof of Theorem 3 is easier to follow when the result is considered as a statement about edge-colorings of “complete graphs with loops,” and secondly, Theorem 3 implies a result about edge-colorings of complete graphs with prescribed colors on some edges, which to us seems almost as interesting and important as the main theorem itself.

An edge-coloring of a graph G is an assignment of a color to each edge and loop of G in such a way that all edges and loops incident with the same vertex have distinct colors. The chromatic index x’(G) of G is the least integer k for which G has an edge- coloring with k colors (an edge-coloring as defined here is sometimes called a proper edge-coloring, and the chromatic index is also called the edge-chromatic number).

The chromatic index of the complete bipartite graph Kn,fl and the complete graph K, is well-known.

n - 1 if n is even, if n is odd . = [ n


We now let KA denote the graph obtained from K, by adding one loop at each vertex. An easy consequence of Proposition 5 is:

Proposition 6.

It is worth noting that in an edge-coloring of KL with n colors, every color occurs at every vertex. This also holds for edge-colorings of K, , , with n colors and of K,, n even, with n - 1 colors, whereas in an edge-coloring of K,, n odd, with n colors each color is missing from exactly one vertex.

We shall not refer much to Propositions 5 and 6, but they explain why we only consider edge-colorings of K , when n is even with n - 1 colors, and of K , when n is odd, K, ,, and KA with n colors. Such edge-colorings are closely related to latin squares. We describe briefly this well-known relationship.

Let L = ( l i j ) be a latin square of side n . Let the vertex classes of K , ,, be {al , . . . ,a,} and (61,. . . , b,}. Define an edge-coloring of K,,, by letting the color of the edge aibj be lij (for all i , j ) . This clearly gives an edge-coloring with ,y(K,,,) = n colors, and it is obvious that the construction also works the other way round and that it works for partial latin squares and partial edge-colorings as well. Latin squares and edge-colorings of K,,n with n colors are equivalent concepts. Thus we can reformulate Theorem 2:

Theorem 7. Let n 2 1. Let G be a spanning subgraph of K, ,, with at most n + 1 edges. Then an edge-coloring of G with at most n colors can be extended to an edge- coloring of K , ,, with n colors if and only if it is neither of Type 1 or 213 of Figure 4, possibly with a jiirtlzer edge, nor of any of Types 415, 6, 718, 9, lO/ll, 12, or 13.

We note that, in Theorem 7, it is convenient, but no restriction, to consider only spanning subgraphs of K, ,n . Any subgraph can be made spanning by adding isolated vertices. We shall continue to make this kind of assumption in the following.

Now consider a symmetric latin square S = (sij) of side n. Naturally it gives rise to an edge-coloring of K,,, in the way just described, but it also gives an edge-coloring of K t , and it is here that our main interest lies. Let the vertices of Kfi be vl, . . . , v,. Define the color of the edge vivj to be sij, and the color of the loop on vi to be sii, i # j , 1 5 i 5 n , 1 5 j 5 n. This clearly defines an edge-coloring of KA with n colors, and it obviously works the other way round as well; it also works for partial symmetric latin squares and edge-colored subgraphs of Kf,. In this connection we note that if an edge- colored subgraph G of Kfi corresponds to a partial symmetric latin square P , then the number of filled cells of P is equal to the degree sum xy=l d(vi) = 21E(G)I + IL(C)l, where E(G) is the edge-set of G and L(G) the loop-set of G (a loop is not an edge: a loop on vi contributes one to the degree d(vi) of vi). We shall say that G has an admissible loop-coloring if the diagonal of P is admissible.

In this terminology, Theorem 3 becomes:

Theorem 8. loop-coloring. Let t be the degree sum of G.

with n colors.

Let n 2 1 and let G be an edge-colored subgraph of Kf, with an admissible

If t 5 n - 1, then the edge-coloring of G can be extended to an edge-coloring of KA

EVANS CONJECTURE t 2 .

0 % I-'

X 7 0

Type 1 (1s x < n)

q n - 2

0 0

Type 7/8 (n 2 4)

2 .

-+' X

x + 1 - x + 1 o-----o

0

"R1 17-2

0 0 x: 1 - Type 2/3 (1s x i n) Type 4/5 (n 2 3)

9' L n - 3

1 - o l o

2 5

0 0 0 0 2 0

Type 9 (n Z 4) Type 10/11 (n 2 5) Type 12 (n t 5)

203

1 - 1 - 1

2

0 0

Type 6 (n 2 3)

2 1

3

4 - Type13(n=4)

Fig. 4. Noncompletable Edge-Colored Spanning Subgraphs of Z& with at most n + 1 Edges.

I f t = n , then the edge-coloring of G can be extended to an edge-coloring of K,!, with n colors if and only if G is not of any of the Types e l , 01 or 02 of Figure 5.

If t = n + 1, then the edge-coloring of G can he extended to an edge-coloring of Kf, with n colors if and only if G is neither of any of the Types e l , 01 or 02, with a further loop added, nor of any of the Types e2, e3, 03, 5a or 5b of Figure 5.

Theorems 3 and 8 are equivalent, and it is Theorem 8 that we shall prove in this article. This will be done in Section 7. Theorem 8 implies a result about completing edge-colorings of complete graphs without loops. As indicated in Proposition 5, we color the edges of K,, with n - 1 colors if n is even, and with n colors if n is odd. Therefore we state the result as two separate theorems.

Theorem 9 is about complete graphs of even order. The odd order case, stated as Theorem 10 below, might seem to be a stronger result; for if m edges are precolored, we can always complete to an edge-coloring of K2,,-1, whereas, in the even case, we only have that if m - 1 edges are precolored, then we can complete to an edge-coloring of Kzm. The reason for this is that we use 2m - 1 colors in both cases and thus in


n even:

Type e l 8 - 0 . 8 - odd number of 1s

Type e2 r... 8l

n odd:

Type 01 8 * * - 8 - Type 5b 1 8 2 _ 1 Q2

t = n

F t = n + l

00 3 o r 4

t = n

t = n

t = n + l

t = n + l

t = n + l

n > 4

n 2 4

1-124

n t 3

n 2 3

n 2 5

n = 5

n = 5

Fig. 5. Noncompletable Edge-Colored Spanning Subgraphs of K j with Degree Sum at Most m + 1 Edges.

the odd case we gain the freedom of having each color missing at some vertex. In fact Theorem 10 is an easy consequence of Theorem 9, as we show below.

We shall refer to the partial latin squares and edge-colored graphs of Figures 1-5, and also those of Figures 6-8 to follow, as bad cases.

Theorem 9. Let G be an edge-colored subgraph of Kzm. For m 2 4 we have:

(a) If IE(G)I 5 rn - 1, then the edge-coloring of G can be extended to an edge- coloring of K2,, with 2rn - 1 colors.

(b) If IE(G)I = m, then the edge-coloring ofG calt be extended to an edge-coloring of Kzm with 2m - 1 colors if and only if G is not of Type 1 of Figure 6.

(c) If IE(G)I = m + 1, then the edge-coloring of G can be extended to an edge- coloring of K2m with 2m - 1 colors if and only if G is neither of Type 1 with an edge added, nor of any of Types 2 or 3 or Figure 6.

For rn = 3: (a) and (b) hold, but if IE(G)I = m + 1 = 4, then the edge-coloring of

EVANS CONJECTURE

Type 1

Type 2

Type 3

Type 6i

Type 6ii

205

m edges

m + 1 edges

rn + 1 edges

m + l =4edges 0 0

- 0 3 - 4 rn + 1 = 4 edges

Fig. 6. Noncumpletable Edge-Colored Spanning Subgraphs of KZm with at most rn + 1 Edges.

G can be extended to an edge-coloring of K6 with 5 colors if and only if G is neither of Type 1 with and edge added nor of any of Types 2, 3, 6i or 6ii of Figure 6.

From m = 2: The edge-coloring of G can be extended to an edge-coloring of K4 with 3 colors if and only if G does aot contain a subgraph of Type 1 .

ProoJ The statement for m = 2 is obvious. Now let G be an edge-colored subgraph of Kzm with IE(G)l 5 m + 1. Let v be a vertex of G of maximum degree, and let G' be obtained from G by replacing each edge incident with v by a loop of the same color on the other end-vertex, and deleting v. Then G' Kim-.l, the degree sum of G' is 2)E(G)I - d ( v ) 5 2m, and G' contains a bad case (with respect to completion to Kf,,-l) if and only if G does (with respect to Kzm) (if G contains Type 1, 2, 3, 6; or 6ii, respectively, of Figure 6 then G' contains Type (01 or 02), 01, 03, 5b or 5a, respectively, of Fig. 5). Clearly, the loop-coloring of G' is admissible. By Theorem 8, G' can be completed to an edge-coloring of Kkmp1 if and only if G does not contain a bad case. But clearly completing G' to an edge-coloring of K i m - , with 2m - 1 colors is equivalent to completing G to an edge-coloring of K2m with 2m - 1 colors, loops in Kkm-l corresponding to edges incident with v in Kzm. Theorem 9 follows immediately. 0

Theorem 10. Let G be an edge-colored subgraph of Kzm- 1.

For m 2 4 we have:

(a) If IE(G)I 5 m, then the edge-coloring of G can be extended to an edge-coloring of Kzm-l with 2m - 1 colors.

(b) I f IE(G)I = m + 1, then the edge-coloring of G can be extended to un edge- coloring of Kzm-1 with 2m - 1 colors ifand only if G is not of Type 4 ofFigure 7.

For m = 3: (a) holds, but if IE(G)I = m + 1 = 4, then the edge-coloring of G can

206

Type4 1 1 * a 11 2 b

4

ANDERSEN AND HILTON

Type5 10 o

Fig. 7. The Noncompletable Edge-Colored Spanning Subgraphs of with m + 1 Edgcs.

be extended to an edge-coloring of K6 with 5 colors if and only if G is not of Types 4 or 5 of Figure 7.

Proof: Let G be as in Theorem 10 with lE(G)I 5 m + 1. Add an isolated vertex to G to obtain G’ G Kzm. Clearly an edge-coloring of Kzm with 2m - 1 colors gives an edge-coloring of K2,,-1 with 2m - 1 colors (the induced coloring), but the converse also holds: for any edge-coloring of Kzrn-l with 2m - 1 colors, each edge joining the new vertex to a vertex of K2rn-1 can be colored by the color missing at this vertex. Now G+ does not contain any of Q p e s 1, 2 or 6ii because these do not have isolated vertices. Hence, by Theorem 9, the edge-coloring can be extended if and only if G + is not of q p e 3 or Type 6i, which corresponds to G not being of Type 4 or Type 5. This proves Theorem 10. 0

Thus Theorem 9 implies Theorem 10. We wish to make a few remarks about the relationship between Theorems 8 and 9. It is certainly possible to write down a direct proof of Theorem 9 with no apparent connection with symmetric latin squares, and we have such a proof which is considerably shorter than this article (although not short in absolute terms). So, even though we consider Theorems 9 and 10 to be very important, we definitely consider the solution of the “symmetric Evans conjecture” (Theorem 3) to be the main feature of this article. Two reasons can be adduced why the proof of Theorem 3 (or 8) should be expected to be more complicated than that of Theorem 9. Firstly, Theorem 9 and so also Theorem 10 follow from Theorem 8 for odd n only. And secondly, Theorem 9 is actually weaker than Theorem 8 for odd n. In Theorem 11 below we state a strengthening of Theorem 9 which is equivalent to Theorem 8 for odd n, and in Theorem 12 we state the similar result corresponding to Theorem 10. Theorem 11 follows from Theorem 8 exactly as Theorem 9 does, and Theorem 12 is easily deduced from Theorem 11. We let A(G) denote the maximum degree of a graph G.

Theorem 11. Let m 3 3, and let G be an edge-colored subgraph of Kzm for which 21E(G)I - A(G) 5 2m. The edge-coloring can be completed to an edge-coloring of Kzm with 2m - 1 colors if and only i f G does not contain Type 3 of Figure 6 or Type 2e of Figure 8 for m 2 4; if m = 3 then G must not contain Types 6i or 6ii of Figure 6 either.

Type 2e

X

2 s x 5 2m - 2

Fig. 8. A(C) = 2m - 1.

A Noncompletable Edge-Colorcd Spanning Subgraph G of Kzm satisfying 21E(G)I -


Theorem 12. Let m 2 3, and let G be an edge-colored suhgraph of Kzrn-l for which 21E(G)] - A(G) 5 2m. The edge-coloring can be completed to an edge-coloring of Kzrn-l with 2m - 1 colors if and only if G does not contain Type 4 of Figure 7 for m 2 4; if m = 3 then G must not contain Type 5 of Figure 7 either.

3. SOME DEFINITIONS AND A BRIEF OUTLINE OF THE PROOF OF THEOREM 8

The proof of Theorem 8 will take up the largest part of the remainder of this article. It will be by induction on n; below we describe how the proof proceeds. First a few definitions.

An edge of a graph G is said to be free if neither end-vertex is incident with any other edge or loop of G, and a loop is said to be free if its vertex is not incident with any edge (our graphs have at most one loop on each vertex). Two edges are independent of each other if they do not have an end-vertex in common; thus a free edge is independent of any other edge of the graph. A set of painvise independent edges is called a matching, and a matching having an edge incident with each vertex is called a 1-factor.

Given a graph G, we let G f denote the graph obtained from G by deleting all free edges and their end-vertices, as well as all isolated vertices, and we let GI denote the graph obtained from G by deleting all free loops and their vertices together with all isolated vertices.

The induction step in the proof of Theorem 8 is basically (but not in all cases) as follows: Given G , we delete an edge e of Gf or G I , i.e., a preassignment. By the induction hypothesis, we extend the edge-coloring of G - e to an edge-coloring of KLP2 with n - 2 colors. We then consider the complete graph with loops K t C KL-2 on the vertex set of either Gf or Gl. This K: has an edge-coloring induced by the edge- coloring of Kfl-2, and we modify the edge-coloring of K: to achieve two things: the edge e must get its prescribed color, and two new colors must be introduced on a number of edges of Kf. This will enable us to extend the modified edge-coloring of Kt to the desired edge-coloring of KA with n colors, making sure that the free edges or free loops get the correct colors.

For even n the above argument is used on GI only. For add n, which of the two graphs Gf and GI we work with is determined by their numbers of vertices, and we define G, to be whichever of Gf and GI has fewest vertices; if they have the same number of vertices, let G , = G1.

There are situations where the above line of attack does not work; for example, the loop-coloring of G might not be admissible with respect to completing to an edge-coloring of KL-2, and such cases will be dealt with separately. Likewise, some other observations will be invoked to facilitate the implementation of the method. But the description above gives the main idea of the proof.

For this plan to be carried out, we need preliminary results of three types. In the next section we discuss the first of these, giving necessary and sufficient conditions for an edge-coloring of Kf to be extendable to an edge-coloring of Kf, with either some free loops or some free edges outside K: getting prescribed colors (in the case of free edges we have to require that at least one vertex of Ki - K,! is not incident with a free edge). In relation to the proof of Theorem 8, this will answer the following two questions: What do we know about the edge-coloring of K:, bearing in mind that it is induced by an edge-coloring of KL-2? And what must we require about the modified edge-coloring of K: for it to be embeddable in Iii?


Then, in Section 5 , we discuss a lemma stating that the modification needed can be brought about. It says that we can find a set of disjoint paths in K: which can be used to make two new colors occur on the required number of edges or loops of KL, while still permitting the old colors to occur sufficiently often; clearly the chromatic index of a graph consisting of disjoint paths is at most two. For the lemma to work, there is a bound on r in terms of n, and in Section 6 we prove some results about the number of vertices of a graph such as Gj or GI. These results show that r = IV(G,)l satisfies, or comes close to satisfying, the required conditions.

The actual proof of Theorem 8 is in Section 7.

4. EXTENDING EDCE-COLORINGS OF Kf WITH FREE EDGES OR FREE LOOPS TO Kf,

Many results have been proved about completing partial latin squares to latin squares, allowing the completcd square to be somewhat bigger than the partial one. Equivalently, edge-colorings of graphs have been extended to edge-colorings of complete bipartite graphs, or of complete graphs, with more vertices.

Much attention has also been given to situations where a particular set of cells outside the partial latin square may also be prescribed, or to the corresponding graph theoretic situations. A typical example is where the diagonal of the latin square is prescribed. We shall use the following such result, proved independently by Hoffman 1171 and the first author [l]. Other results of this type can be found in [4], [8], [9], [14], [15], 1161, [lS], [19], and [20]. If a color ck occurs on exactly e edges and exactly 1 loops of a graph G, we say that ck occurs with degree sum s = 2e + 1 in G .

Theorem 13 [l], [17]. Let G be a subgraph of K,!,, consisting of a Kf and possibly some free loops. Let G have an edge-coloring with n colors e l , . . . , cn, for each k let l ( k ) denote the number offree loops of color ck, and let the loop-coloring be admissible. Then the edge-coloring can be completed to an edge-coloring of KL with the same colors if and only iJ for 1 5 k 5 n, the color ck occurs with degree sum at least 2r - n + E(k) in the K:.

In [7], we proved the following theorem about the case with free edges:

Theorem 14 [V. Let G be a spanning subgraph of K;, consisting of a K:, at least one isolated vertex-, andpossibly somefree edges. Let G have an edge-coloring with n colors c1,. , . , c,, for each k let e (k ) be the number of free edges of color ck, and let the loop- coloring be admissible. Then the edge-coloring can be completed to an edge-coloring of KL with the same colors if and only if, for 1 5 k I n , the color ck occurs with degree sum at least 2r - n + 2e(k) in the K:.

Both Theorem 13 and Theorem 14 are generalizations of a well-known result of Cruse [lo] (obtained by putting all E(k) or all e ( k ) equal to zero).

The condition in Theorem 14, that at least one vertex of Kf, should not be incident with and edge or loop of G, is necessary (see [7]).

In [7], a stronger version of Theorem 14 is proved, where the number of loops of each color can be specified. This version actually includes Theorem 13 as a special case. In general, as noted in 171, the case where both free edges and free loops are prescribed


seems to be a good deal more difficult, with no simple necessary and sufficient condition such as in the above theorems. However when n is even, in one particular situation we do have a result with free edges and free loops. The proof uses a mild form of a trick we use in Section 7.

Theorem 15. Let n be even, and let G be a subgraph of K i consisting of a Kf , possibly some free edges, and possibly some free loops. Let G have an edge-coloring with n colors c1, . . . , c,,, and for 1 5 k I n, let s (k ) be the number of loops of the K: colored ck, let e (k ) be the number of free edges colored ck, and let l ( k ) be the number offree loops of color ck; put l * ( k ) = s (k) + 2e(k) 4- l (k) .

I f ( l* ( l ) , . , , , l*(n)) is the sequence of an admissible loop-coloring of a Kf,, then the edge-coloring of G can be completed to an edge-coloring of Kfi with the same colors if and only if, for 1 5 k 5 n, the color c k occurs with degree sum at least 2r - n + 2e(k) + l ( k ) in the K f .

Proof. Suppose first that the edge-coloring is completable, and consider any k, 1 5 k 5 n. In K i - K f , the color ck must occur at at least 2e(k) + E(k) vertices, so it can occur on at most (n - r ) - (2e(k) + l (k ) ) edges joining a vertex of Kf, - Kf to a vertex of K f . Therefore, in K f , it must occur already at at least r - ( n - r - 2e(k) - l ( k ) ) vertices. This proves the necessity of the condition.

To prove the sufficiency, suppose that each ck occurs the required number of times. Let G* be the graph obtained from G by replacing each free edge by a loop on each of its end-vertices, giving both loops the color previously on the edge. In G*, each color ck occurs on 2e(k) + l (k ) free loops (and on l* (k) loops altogether). It follows from the assumption that the loop-coloring of G" is admissible, and by Theorem 13 the edge-coloring of G" can be completed to an edge-coloring of Kf,. In this edge-coloring of K i , we interchange, for each free edge of G, the color of the edge and the common color of its end-vertices. Clearly this again yields an edge-coloring of KL, and it clearly extends the original edge-coloring of G.

0 This completes the proof of Theorem 15.

In Theorem 15, if (Z*(l), , . . , l*(n)) is not admissible but all the other conditions are

We also need three corollaries: satisfied, it may or may not be possible to complete the edge-coloring.

Corollary 16. Let G be a subgraph of Kf, with no edges, and let the loops of G be colored with an admissible loop-coloring. Then this can be completed to an edge-coloring of Kfi.

Prooj We apply Theorem 13 with r = 1. The numerical condition is vacuous unless Z(k) = n - I , in which case n is even (or n = 1) and no color different from c k can occur on any loop of G; then the condition is satisfied with ck on the loop of the K1. 0

Corollary 17. Let Kt KL, r 5 ; + 1, and let all loops of Kf, be colored so as to give an admissible loop-coloring. If some edges of Kf are colored subject to the provisos that at most n colors are used altogether, and each color c occurs on at least 2r - n + l (c ) vertices of K f , where l (c ) is the number of loops outside Kf. of color c , and in the case r = 5 + 1 there is some color which occurs on each vertex of Kf , then the edge-coloring can be completed to an edge-coloring of Kf, with n colors.

21 0 ANDERSEN AND HILTON

ProoJ We can color all uncolored edges of K: one by one, because the number of colors already occurring at each end-vertex is at most r - 1, so the number of forbidden colors is at most 2(r - 1 ) 5 2(; + 1 - 1) = n, and if there is equality here, then r = 5 + 1 and some forbidden color is counted twice; thus there is always at least one color available.

The assumptions imply that each color occurs at the number of vertices required by 0 Theorem 13, so we can use that theorem to complete the edge-coloring.

Corollary 18. KL, r 5 q. I f some edges and loops of K: are colored subject to the provisos that at most n colors are used altogether, all colored loops have distinct colors, and if r = 9 then each color occurs at least once, then the edge-coloring can be completed to an edge-coloring of KL with n colors.

Let n he odd, und let KL

Prooj We can first color all uncolored loops of Kf one by one, admissibly, because the number of colors already occurring at a vertex with uncolored loop is at most r - 1, and the number of loops colored so far is at most r - 1, so the number of forbidden colors is at most 2(r - 1) 5 2 = n - 1, so there is always at least one color available. In the same way, we can color all uncolored edges of K:. We then apply Theorem 13 with

0 all l (k) equal to zero.

5. PATH SYSTEMS WITH ALL EDGES HAVING DISTINCT COLORS

In this section we prove some results enabling us to find, in an edge-colored K,, a sufficiently large subgraph whose edges can be recolored with two new colors, which does not contain any of a given set of edges (those with prescribed colors) except, if required, one or two that so far have incorrect colors.

First some terminology. By a path system of a graph we simply mean a subgraph consisting of disjoint paths. Tf F and M are two sets of edges with F f l M = 0 ( F for forbidden, M for mandatory), a path system whose edge set includes every edge of M but no edge of F will be called an ( M , F)-path system. If all its edges have distinct colors, we call it an ( M , F)-d-system for brevity. Such a system having exactly i edges, we call an i - ( M , F)-d-system.

A path with j edges we call a j-path. Similarly for a j-circuit and j-star. Finally, the graph with edge-set F and with no isolated vertices will be denoted by G[F].

Our tool in this area is the following theorem due to the first author 131.

Theorem 19. Let r 2 2, and let Ki have a (proper) edge-coloring with any number of colors. Let F be a spanning edge-set of K,, and let M be the edges of a path system of K,, where F n M = 0 and all edges of M have distinct colors. Put

& ( I , F , M ) =

1 - ( J r z - (22 - 41Ml)r + 961FI + 4(IMI2 + 1Ml) + 9 - r + 21MI + 5 ) . 8

Then K , contains an ( M , F)-d-system P with

EVANS CONJECTURE 21 1

Note that since F is a spanning edge-set of K,, IF1 2 i, so 96)FI - 22r 2 26r, and

In our applications of Theorem 19, IMI will always be 0, 1 or 2. so E ( r , F , M ) is always a real number.

Corollary 20. Let K , have an edge-coloring with any number of colors, and let F be a spanning set of edges of K , . Then K , has an

(1) (0, F)-d-system with at least 2 r - n edges, for all integers n 2 4, all integers r 5 7, and all F with IF1 5 ;.

(2) (0,F)-d-system with at least 2r - n edges, for all integers n 5 6 , all integers r 5 7, and all F with IF1 5 7.

(3) ( M , F)-d-system with at least 2r - n edges, for all integers n 2 1, all integers r 5 7, all M with (MI = 1 and all F with IF1 5 F.

(4) ( M , F)-d-system with at least 2r - n edges, for all integers n 2 1, all integers r 5 F, all M with (MI = 2, the two edges having distinct colors, and all F with IF1 5 7.

3n-1

311-3 n + l

3n-4

R +4

Proof: We shall prove the corollary by an application of Theorem 19, but as this does not work for small n, we first prove the existence of certain path systems directly. In a path system, we shall refer to the components with edges as path components. When IMI = 2, we shall always assume that the edges of M have distinct colors, and F is always spanning.

a. K6 contains a 3-(M, F)-d-system, for all M with IM( 5 1 and F with IF1 5 8.

Proof. We may assume that (MI = 1, say e E M . If there is an edge e’ independent of e and not in F of color different from the color of e , then the system containing e and e’ can be extended with a third edge, because of the 15 edges of K6 at most 6 are forbidden because their color is used (including e and e‘), and at most 8 are forbidden because they belong to F. If such an edge e’ does not exist, then there are at least 4 edges of F not incident with the end-vertices of e , and so an edge can be added incident with each end-vertex of e , to form a 3-path.

b. KI contains a 3-(M, F)-d-system, for all M with IMI 5 2 and F with IF1 5 9.

Proof: We may assume that IMI = 2. There is a third edge to add to M, because of the 21 edges of K7 at most 6 are forbidden because their color is used (including the two belonging to M ) , at most 9 are forbidden because they belong to F , at most a further 3 because they are incident with a vertex already incident with both edges of M (if M is a 2-path, one such edge is in F), and at most 1 because it forms a circuit with the edges of M .

c. K7 contains a 4 - ( 0 , F)-d-system, for all F with IF1 I 5.

Proof: Since F is spanning, G [ F ] must have at least 2 components, and one of them must have either 2 or 3 vertices. It follows that the complement contains K2,5 or K3,4; either of these will contain the required path system.


d. KS contains a 4-(M, F)-d-system, for all M with IMI 5 2 and F with IF1 5 9.

Proof We may assume that 1M1 = 2, say M = { e . e’}. If e and e’ are independent, a third edge can be added so that the 3 edges do not form a 3-path, because of the 28 edges at most 8 are forbidden because of color, at most 9 because they belong to F , and at most 4 because they would form a 3-path with e and e‘. If e and el form a 2-path, we also extend so as not to create a 3-path, if possible. If this cannot be done, then at least 6 edges of F must join pairs of vertices not incident with e or e’ (of the 10 such edges only 4 can be forbidden because of color); but in that case we can extend directly to a 4-path. So we assume that we have a path system with 3 edges which is not a 3-path. Then we can add a fourth edge, because the number of forbidden edges is at most 12 due to color, 9 due to F , 4 more due to a vertex of degree 2 in the system, and 1 due to the creation of a triangle.

e. Kg contains a 5 - ( 0 , F)-d-system, for all F with IF1 5 5 .

Prooj we prove that either of these will contain the required path system.

G[F] must have at least 3 components, the complement contains K3,5 or K2,2,4;

Suppose the complement contains a K3,5. Let U be the set of 3 vertices of degree 5, and let V be the set of 5 vertices of degree 3. Let the vertices of U be u1, u2, u3 and of V be v1, v2, vg, v4, v5. It is easy to see that there is a path of 4 distinctly colored edges, say V I U I V ~ U ~ V ~ colored 1, 2, 3 ,4 in order. There is an edge colored with a further color, say 5, incident with 143, and we have a 5 - ( 0 , F)-d-system unless ~ 3 ~ 2 is colored 5, and the other edges incident with u2 have colors 1, 2, 3, 4. In that case, we have the desired system unless the color 2 is on u3v3, and similarly we have the system unless V I U ~ is colored 3. So suppose ~ 3 ~ 2 , u3v3 and ~ 3 ~ 1 are colored 5, 2, and 3, respectively, and that u3v4 has color 1 and 4 1 1 5 color 4. By the same argument applied to the path ~ 4 ~ 3 ~ 3 ~ 2 ~ 2 (colored 1,2,4,3) we have a 5 - ( 0 , F)-d-system unless ul v3 is also colored 5 (or a new color, say 6). But in that case we get a 5-(0,F)-d-system consisting of the path ~ 5 ~ 3 ~ 3 ~ 1 ~ 1 (colored 4,2,5(6), 1) and the edge u2v2 (colored 3).

Now suppose the complement contains K2,2,4. Let the independent vertex-sets be U = {UI, U Z } , V = (v1, v2) and W = {w,, w2, w3, wq}. We may suppose that the edges ulvl and u1v2 are colored 1 and 2, respectively. There are at least two edges joining u2 to W of color different from 1 and 2, so we may suppose that u2w1 and u2w2 are colored 3 and 4, respectively. The colors used on the four edges joining v2 to W include a further color, say 5. The edge incident with v2 of color 5, together with uIv1, u1v2, u2wl and uzw2 form the required path system.

f. Kq contains a 4-(M, F)-d-system, for all M with IMI 5 2 and F with (FI 5 9.

Prooj necessary adding one or more edges in the resulting Kg to F to obtain a spanning set.

Follows from d by deleting a vertex not incident with an edge of M , and if

g. Kq contains a 5-(0, F)-d-system, for all F with IF1 5 7 .

Proof. We may assume that IF1 = 7. G[F] has a vertex v either having degree at least 3 and whose component is not a 3-star, or having degree 2 with both neighbors


of degree 2. Deleting v and adding the minimal number of edges necessary to ensure that the graph obtained spans Kg gives a spanning subgraph of Kg with at most 5 edges. The results now follows from e.

h. Klo contains a 5 - (M, F)-d-system, for all M with IMI 5 2 and F with IF1 9 11.

ProoJ We may assume that lMl = 2 and IF( = 11. Then G[F] has a circuit containing a vertex v not incident with any edge of M . We delete v, and let F' be a spanning set of edges of KS obtained from G[F] - v by adding edges incident with isolated vertices, if needed; since at least two neighbors of v are not isolated, IF'I 5 9. By f, KS has a 4-(M, F')-d-system. So our Klo has a 4-(M, F)-d-system; we first prove that we can assume that it is not a 4-path. So suppose that it is, and let the colors of the edges be, successively, 1, 2, 3, and 4. If M consists of the edges of color 2 and 3, consider the K7 spanned by the vertices not incident with any edge of M . If both colors 1 and 4 occur on nonforbidden edges in this K7, we get a 4-(M,F)-d-system which is not a 4-path by including two such edges in place o f the edges of the 4-path of colors 1 and 4, so suppose that 1 does not occur. We try to replace the edge of color 4 by an edge in the K6 avoiding the first 4 vertices of the 4-path: out of 15 edges, at most 6 may be forbidden because of colors 2 and 3 (and none because of color l), and at most 9 because they belong to F; if there is equality in both cases, then no edge joining the first vertex of the path to the K6 is in F , and we can instead replace the first edge by an edge joining this vertex to a vertex not in the 4-path. In any case we obtain a 4-(M, F)-d-system which is not a 4-path. If, say, the edge of color 3 is not in M , we replace it by an edge in the K7 spanned all vertices except those of degree 2 in the 4-path; of the 21 edges at most 9 are forbidden due to colors 1, 2, and 4, at most 9 due to F, and 1 because it would give a 4-path. So we can indeed assume that the system of 4 edges is not a 4-path.

Counting the number of edges which can be colored 1, 2, 3, or 4 and the numbers of forbidden edges (those in F and those giving circuits) in the complete graph obtained by deleting vertices of degree 2 in the nonpath 4-(M, F)-d-system (of which there are at most 2) then shows that the system of 4 edges can be extended with a fifth edge.

i. Klo contains a 6 - ( 0 , F)-d-system, for all F with IF1 5 7

Proof. We may assume that IF1 = 7. Deleting a vertex v of degree 1 in G[F] and using g, we obtain a path system P with 5 edges in Klo - v. The number of vertices in Klo - v that are either end-vertices of path components or isolated vertices with respect to P is at least 5 , and if v is joined to such a vertex by an edge which is not in F and whose color is not used in P , then P can be extended by adding this edge. But v is incident with one edge of F, and at most 5 edges whose color is used in P , so it is incident with at least 3 usable edges. If they all go to intermediate vertices of P, then v is joined to two successive vertices u and u' of a component by usable edges, and so a larger path system can be obtained by excluding uu' and including vu and vu'.

j. K11 contains a 6 - ( M , F)-d-system, for all M with ]MI 5 2 and F with IF1 5 11.

Proot We may assume that IMI = 2 and IF1 = 11. Some vertex v not incident with any edge of M has degree at most 2 in G[F]. It follows from h that there is a 5 4 4 , F)-


d-path system in K11 - v. Suppose first that this is a 5-path. Then this can be extended, unless the 6 vertices of Klo that do not have degree 2 in the 5-path are all joined to v by forbidden edges or edges with used colors. But then at least 3 of the remaining vertices must be joined to v by usable edges, at least two of them consecutive. The edge between two such consecutive vertices can be replaced by two edges to v, unless it is in M. If there are 4 such consecutive vertices we can extend the path system, so we may now suppose that there are no such 4 consecutive vertices. Then v must be incident with exactly 2 edges of F , an edge not in F of each color of the 5-path, and exactly 3 edges not in F of other colors. It follows that each of at least two edges of the 5-path can be replaced by a usable edge joining an end-vertex of the edge to v , and then, possibly, another edge having the same color as the omitted edge can be added. If this is not the case, each of these two colors occurs on at most one edge (joining v to the 5-path) in the K7 spanned by the 7 vertices not having degree 2 in the original 5-path. Then the number of forbidden edges in K7 for extending the original 5-path is at most 2 + 3 X 3 = 11 due to color and 9 due to F; even if the only available edge joins the end-vertices of the 5-path, we can extend.

If the 5-(M,F)-d-system has 2 components the same argument, in a simpler form, works. And if the system has at least 3 components, it works in a rather trivial way.

k.

Proof. We may assume that IF1 = 7. Deleting a vertex v of degree 1 in G[F] and using i, we obtain a path system P with 6 edges in K11 - v. At least 3 edges incident with v are not in F and are not colored with a color used in P ; if one of these edges joins v to an end-vertex of a path component of P or an isolated vertex with respect to P, or if two of them go to consecutive vertices of P , we get a 7-(0, F)-d-system. One of these will occur, unless we have one of three situations: P is a 6-path, the usable edges go to the second, fourth, and sixth vertex; P is a 4-path and a 2-path, the usable edges go to the second and fourth vertex of the 4-path and the middle vertex of the 2-path; or P is three 2-paths, the usable edges go to the middle vertices. In all three cases, if e is a nonforbidden edge joining v to a vertex not in P , there is a 7-(0, F)-d-system containing e , a usable edge incident with v, and the edges of P except the one having the color of e . By counting we see that there is such an edge e unless P consists of three 2-paths, and the only vertex u of Klo not in P is joined to v by an edge of F. But since F is spanning and IF1 = 7, not all edges joining u to the end-vertices of the 2-paths of P can be in F, and so we can exchange such an edge with an edge of P and avoid this situation.

K11 contains a 7-(0, F)-d-system, for all F with IF1 5 7.

1. K12 contains a 6-(M, F)-d-system, for all M with IMI 5 2 and F with IF1 5 12.

Pro08 Let v be a vertex of maximum degree in G[F] among those not incident with any edge of M. Let F' be a spanning edge-set of K12 - v obtained from the edges of F not incident with v by adding as few edges as possible to make the edge-set spanning. We show that we can choose v so that IF'I 5 11; the statement then follows from j. If IF1 I 11, then clearly IF'I I 11, so assume IF1 = 12. If v has degree 1 in G[F], then since it has maximum degree, at most 12 - IV(G[M])I edges of F are incident with a vertex outside G[M], so IV(G[M])( edges of F join two vertices of G [ M ] ; it follows that


IV(G[M])I = 4 and that the neighbor of v is in G [ M ] and incident with another edge of F , so that no edge has to be added to obtain F'. So in this case, IF'I = 11. If v has degree 2 in G [ F ] , clearly IF'I 5 11 except if the two neighbors of v are joined by an edge of M and both incident with no other edge of F ; but then the remaining 10 edges of F would join vertices among the remaining 9 vertices, and so another choice for v must exist, for which IF'I 5 11. Finally, if v has degree at least 3 in G [ F ] , there is no problem.

m. K13 contains a 7 - (M, F)-d-system, for all M with IMI 5 2 and F with IF1 5 12.

ProoJ We may assume that IMI = 2 and IF1 = 12. Let v be a vertex of minimum degree in G [ F ] subject to not being incident with an edge of M . Then v has degree 1 or 2; if it is 2 then at least 7 vertices not incident with an edge of M have degree 2. By deleting v and also a further vertex, we can obtain a K11, containing M , with a spanning set of at most 11 edges, including those of F in K11. Therefore, by j, K13 - v contains a 6 - ( M , F)-d-system P . At most 8 edges incident with v are in F or have a color used already, and if P has at least 3 path components, then K13 - v has at least 9 vertices that are not intermediate vertices of P , so one of them is joined to v by a usable edge, implying that P can be extended.

So assume that P has two path components; if v has degree 1 in G [ F ] we can extend as above, SO assume that the degree is 2, that exactly 4 usable edges are incident with v , and that these join v to the intermediate vertices of P . If P is a 1-path and a 5-path then v is joined by usable edges to both end-vertices for three consecutive edges of the 5-path, and as at most two of these can be in M , we can extend P by omitting one not in M and adding two edges to v. Suppose that P is a 2-path and a 4-path, and let the vertices of the 2-path be, in order, vl,v2, v3. The argument above works unless the two internal edges of the 4-path are in M , so suppose that this is case. Let the color of vlv2 be 1. We can obtain a path system with 7 edges by excluding vlv2 from P and adding vv2 and the edge of color 1 incident with v , except if this edge is vv3. But as some vertex different from v and not in the path components of P must also have degree 2 in G [ F ] , we can exchange v and such a vertex, which cannot be joined to v3 by an edge of color 1 if v is. If P is two 3-paths, a similar argument works.

Suppose finally that P is a 6-path. If v has degree 1 in G [ F ] then it is incident with at least 5 usable edges, and we can extend in one of the two usual ways. The same holds if the degree of v in G [ F ] is 2, except if v has exactly 4 usable edges, joining it to the second, third, fifth, and sixth, or the second, third, fourth, and sixth vertex of P , with each end-vertex of each edge of M being incident with a usable edge. In each of these cases, there will be at least two vertices different from v and not on the 6-path that also have degree 2 in G [ F ] . If either are joined by usable edges to anything other than the same vertices of P as v we can extend, so suppose that they are both joined by usable edges to exactly the same vertices of P . Then omitting the first edge from P , adding the edge joining the second vertex to v and an edge incident with v of the same color as the first edge will work for at least one of the choices for v .

n. K14 contains a 8 4 4 , F)-d-system, for all M with IMI I 2 and F with IF1 5 12.


Proof. We may assume that \MI = 2 and IF1 = 12. Let v be a vertex not incident with an edge of M, such that v has smallest degree in G[F] among such vertices. Then either that degree is 1, or all such vertices have degree 2 and the two edges of M are independent. By m, K14 - v has a 7-(M, F)-d-system P . The number of edges incident with v that are not in F and do not have a color used already is at least 4 if the degree of v in G[F] is 2, and at least 5 if the degree is 1. If such a usable edge joins v to a vertex which is an end-vertex, or an isolated vertex, in P , then this can be extended immediately. Also if two such edges join v to successive vertices of P , and the edge joining these is not in M, then replacing this edge by the two edges incident with v again gives an extension. If P has at least 4 path components, then v will indeed have a usable edge to a nonintermediate vertex, as there will be at least 10 such vertices.

Suppose that P has exactly 3 path components. Again, v is joined to a good vertex by a usable edge, except if v is incident with 2 edges of F and has exactly 4 usable edges, all going to intermediate vertices of P ; in this exceptional case, we get usable edges to successive vertices, with the edge joining them not being in M, unless P consists of either one 1-path and two 3-paths or two 2-paths and one 3-path (P cannot consist of one 1-path, one 2-path, and one 4-path as the edges of M would not be independent then). So suppose that P is one of these types, and that all intermediate edges are in M. Let v1 be the end-vertex of a 3-path, let v2 be the adjacent vertex of the path, where the edge vlv2 is not in M and has color 1. Let the other end-vertex of the 3-path be v‘. Removing vlv2 from P and adding V V ~ , we can obtain an extension by adding the edge of color 1 incident with v, except if this edge is vv’. But since v has degree 2 in G[F] , any one of the vertices not in path components of P can play the role of v, and the obstacle of being joined to v‘ by an edge of color 1 can only happen to one of them.

A similar argument works if P has exactly 2 path components. Finally assume that P is a 7-path. And first suppose that v has degree 1 in G [ F ] .

Then it is incident with at least 5 usable edges, and P can be extended in one of the two usual ways. Suppose then that v is incident with 2 edges of F , and that it has only 4 usable edges. If we cannot extend P in one of the usual ways, then these must go to at least one pair of successive vertices of P , and the edge of P joining these vertices must then be in M. We assume that this happens for all vertices which do not belong to the 7-path (and which could therefore be the vertex v). Then there must be an end-edge e of P , say of color 1, not belonging to M, so that every candidate for v is joined to the end-vertex of e of degree 2 in P by a usable edge; as a matter of fact, except if both the second and fifth edge of P belong to M, all v-candidates must be joined to both “inner” vertices of end-edges by usable edges, and in this exceptional case there are three vertices (the second, third, and fifth) to which they must all be joined by such edges. It follows that in all cases the vertices of the 7-path to which any candidate for v can be joined by an edge of color 1 belong to a given set of at most 5 vertices. But there are 6 candidates for v , so one of them is not joined to the 7-path by an edge of color 1. Then we obtain the required path system by choosing this v, deleting e, adding the usable edge from v to an end-vertex of e, and also adding the edge of color 1 incident with v.

After these preliminary observations, we now prove the four statements of the corollary one by one. The arguments are similar in the four cases, and we only go into details for the first statement.

EVANS CONIECTURE 21 7

1. Let F be a spanning edge-set of cardinality at most 5 , and let M be empty. Assume that 2 5 r 5 1 ?I. We shall prove that both terms (the ceiling in case of the last term, rather) in the minimum of Theorem 19 are at least 2r - n, if n 2 16. First,

51 E r n - - - 3n - 1

3 3 Next, to prove 5 r - &(I, F , 0 ) ] 2 2r - n, we prove 4 r - ~ ( r , F , 0 ) > 2r - n - 1:

1 8

- -(Jr* - 22r + 961Fl + 9 - r + 5)

8 8 9 8

- > n - - r + - - -

Differentiation with respect to r shows that for fixed n, this last expression is a decreasing function of r . For n = 16 we have r I 11, and inserting n = 16, r = 11 gives approximately 0.8; for n = 17 we get r 5 12, and inserting these values gives approximately 0.6; so for n = 16 and n = 17 the expression is greater than zero for all r in our range, as required. Now assume n 2 18. Inserting the upper bound on r gives

9 3 n - 1 3 + -- 8

n--- 8 4

1 9n2 + 1 - 6n 3n - 1 1 + 48n + 9 = - (5n + 21 - J9n2 + 498n + 233). 32

- 22- -/ 8 16 4

The derivative of this last expression with respect to n is

9n + 249 ' 32 ( 5 - J9n2 + 498, + 233

which can be estimated, for n 2 18:


9n + 249 9n + 249

d(3n + 50)' + (198n - 2267) ' 32 ( 5 - J 9 n 2 + 498n + 233

32 3n + 50

32 3n + 50 > 0 .

So the expression is an increasing function of n and hence always at least its value for n = 18, which is approximately 0.03. This proves statement 1 for n 2 18. So far, we have thus proved it for n 2 16.

For smaller values of n statement 1 is either trivial (if 2r - n is negative, 0, 1 or 2) or follows from the small cases done in the beginning in the following way: For n = 15 from k ( r = l l ) , i ( r = 101, and g ( r = 9); for n = 14 from i and g; for n = 13 from g and d; for n = 12 from d; for n = 11 from e and c; for n = 10 from c; and for n = 9 from a. Finally, for n = 7 it is seen directly that KS has a 3-(0,F)-d-system when IF1 = 3.

2. The proof is similar to the above. We get ( r - 5 ) - (2r - n) 2 0 for n 1 14. Since r I 7 and IF1 I 2 we find that 3n -3 n f l

3 1 - r - ~ ( r , F , 0 ) - 2r + n + 1 2 - (5n + 39 - J9n2 + 486n + 1185). 4 32

This expression is positive when n = 14. Differentiating it we obtain an expression which is, for n 2 14, greater than

9n 3n + + 2 4 3 ) * 51 - ( 5 1 -

32

Therefore the statement is true for n 2 14. The nontrivial cases with smaller values of n are covered: For n = 13 by g and d, for n = 12 by d, for n = 11 by b, and for n = 9 by a.

3. In this case ( r - 5 ) - (2r - n ) 2 0 for all n 2 13, but the other inequality needs some care. As IMI = 1 and IF1 I 2, we get

3 9 1 1 - r - & ( r , ~ , M ) - 2 r + n + 1 r n - - r + - - - J r 2 - 1 8 r + 4 8 n + 3 0 5 . 4 8 8 8

As before, for fixed n this is a decreasing function of r , and inserting the values 13, 14, 15, 17, 18, and 19 for n , with the corresponding maximum values 8, 9, 10, 11, 12, and 13 for r we always get something positive, so the statement holds for these values of n; notice that n = 16 is missing from the list.

n+6

For n 2 20, we can insert the bound for r to get

3 1 - r - E ( r , F , M ) - 2r + n + 1 2 - ( 5 n + 40 - J9n2 + 528n + 5184), 4 32

which again is an increasing function of n ; inserting n = 20 shows that it is positive for all n 2 20.

The nontrivial smaller cases again come from the first part of the proof For n = 16 from j and h, for n = 12 from d, and for n = 11 from b.


4. Again, ( r - 5) - (2r - n ) 2 0 for n 2 13. This time, using llwl = 2 and IF( 5 n+4 2 we get

3 9 1 1 - r - &(r ,F,M) - 2r + n + 1 2 n - - r - - - - J r 2 - 14r + 48n + 225. 4 8 8 8

This is a decreasing function of r , and by insertion it is seen to be positive for the values 13, 14, 17, 18, 19, 21, 22, and 23 of n , and the corresponding maximum values 8, 9, 11, 12, 13, 14, 15, 16 of r . We still have to deal with n being 15, 16, and 20.

For n 2 24, however, we can insert the bound for r to obtain

3 1 - r - e ( r , F , M ) - 2r + n + 1 2 - 2 ( 5 n + 32 - J9n2 + 576n + 3840), 4 32

and as this increases with n and is positive for n = 24, the statement also holds for n 2 24.

This again leaves some small cases. We get n = 20 from n, m, and 1, n = 16 from j and h, n = 15 from h and f, n = 12 from d, and n = 1 1 from b.

0 This completes the proof of Corollary 20.

6. SOME LEMMAS ON THE ORDER OF THE DENSE PART OF A GRAPH

In this section we prove some easy lemmas enabling us to bound the order of the graphs on which we wish to apply the results of Section 5.

A component of a graph consisting of one edge with a loop on one of its end-vertices, we shall call a loop-edge pair. We remind the reader that a loop on a vertex contributes one to the degree of the vertex.

Lemma 21. Let G be a graph without isolated vertices and without free loops, and with the property that for each edge there is a loop on at least one of the end-vertices. Let the degree sum of G be s.

Then IV(G)l 5 3 s, with equality if and only if each component of G is a loop-edge pair.

2

Proof. Let G have 1 loops, none of them free, and e edges. Then s = 2e + 1 . If I > e then s > 3e, and as G has no free loops, IV(G)l I 2e, so IV(G)l < So suppose that 1 5 e. Then s 5 3e with equality if and only if e = 1, and as G

contains a vertex for each loop, and at most one further vertex for each edge, we get IV(G)l 5 1 + e = s - e 5 J s, with equality if and only if e = 1 and each edge has one end-vertex of degree 1. The two conditions combine to the requirement that each

0

s.

2

component be a loop-edge pair.

Lemma 22. loops, and suppose that G has e edges and 1 loops.

a loop-edge pair.

Let G be a graph without isolated vertices and without free edges and free

Then IV(G)l 5 y, with equality if and only if each component of G is a 2-path or

Proof. Assume first that G has no loops, and let x be the number of vertices of degree 1. As there are no free edges, x 5 e with equality if and only if each edge has one end-


vertex of degree 1. Now,

2e = 1 d ( v ) 2 x + 2(IV(G)I - x) = 21V(G)I - x 2 2IV(G)l - e , vEV(G)

with equality all way through if and only if x = e and each vertex has degree 1 or 2. From this we get

3 IV(G)l 5 ~e

with equality if and only if all components are 2-paths. Now let 1 > 0. We remove loops successively from G without creating free edges

until the removal of any remaining loop creates a free edge, Let 1’ be the number of loops left. Clearly they all belong to loop-edge pairs, altogether having 2E’ vertices. By the first part of the proof we then have

3e + I’ 3e + 2 IV(G)l 5 - (e - 2’) + 21’ = ___ < - - 3

2 2 2 ’

with equality if and only if 1‘ = 1 and all components not having loops are 2-paths. This proves the lemma. 0

Corollary 23. free loops, and suppose that G has degree sum s and 1 loops.

loop-edge pair.

Let G be a graph without isoiated vertices and without free edges and

Then IV(G)l 5 y, with equal$ if and only if each component is a 2-path or a

Recall that Gf is the graph obtained from G by deleting all free edges, their end- vertices, and all isolated vertices. Similarly GI is the graph obtained from G by deleting all free loops, their vertices, and all isolated vertices. G, is whichever of Gf and Gl have the fewest vertices; if they have the same number of vertices then G, = G I .

Corollary 24. Let G be a graph with degree sum s, with e free edges and 1 p e e loops. Then IV(G,)l 5 7 - max(2e,I}. 3s+2e+l

3 Prooj ( s - 2e - I ) + 1 and IV(Gl)l 5 7 ( S - 2e - 1) + 2e. As IV(G,)l = min{lV(Gf)l, IV(Gl)l} we get IV(Gm)l 5 3 (S - 2e - 1 ) + min{2e,Z} = 5 ( s - 2e - 1) + (2e + Z - max{2e,Z}) = 7 - max{2e,i}. 0

By Corollary 23, IV(Gf)l 5 3

3 3s+2e+I

7. PROOF OF THE MAIN THEOREM

In this section we prove Theorem 8 (and so also Theorem 3, as they are equivalent).

Proof of Theorem 8. We showed in Section 1 that the exceptional cases cannot be completed. It remains to prove that if G is a spanning subgraph of KL and the degree sum of G is at most n + 1, then an edge-coloring of G with admissible loop-coloring, which does not contain any of the Types of Figure 5 , can be extended to an edge-coloring of KL with n colors. We do this by induction on n.

As remarked in Section 2, a graph containing one of the Types of Figure 5, we shall call a bad case. For n = 1 and n = 2 there are no bad cases.

221 EVANS CONJECTURE

n = 4:

n = 5 :

3g$3 2 81 0

0 0 2 ~ 1 fi2

Type e3

Type 5b

3Q--Q404 2 1 0 Type 03

Fig. 9. Small Noncompletable Cases.

We note that by Theorem 4, we have that the theorem is true for graphs G with no isolated vertices. In view of this, it is very easy to check that it is also true for n 5 5; we leave it to the reader to check that if all edge-colored G with an isolated vertex, admissible loop-coloring and degree sum at most n + 1 are examined for n 5 5, the only noncompletable cases encountered are those shown in Figure 9, and these are all bad cases,

We now assume that n 2 6 and that G is a spanning subgraph of Kf, with degree sum at most n + 1, with admissible loop-coloring and with an edge-coloring such that G is a not bad case. We assume that the theorem is true for graphs with fewer than n vertices. We must show that the edge-coloring of G can be completed to an edge-coloring of K!, with n colors (clearly at most n colors are used on G).

We first remark that by Corollary 16, the edge-coloring can be completed if G has no edges, so we assume that this is not the case. Then, if the degree sum of G is less than n + 1, it is possible to add one or more edges and loops to G, with appropriate colors, so as to obtain a graph which is not a bad case and which has degree sum exactly n + 1. An easy way to see this is the following: If the degree sum of G is at most n - 1, we add an edge so that a path of length at least two is in the graph afterwards; as n 2 6, no bad case can be formed in this way. We repeat this until a graph with degree sum n or n + 1 is obtained; if it is n, we add any loop and, if necessary, choose its color so as to avoid the bad cases; again this is easily achieved. So we assume in the following:

G has degree sum exactly n + 1. (1)

Let us remember that by Theorem 4, we also may assume:

G has an isolated vertex. (2)

For the remainder of the proof, we consider the cases of even and odd n separately. Although there are quite a few similarities in the proofs for the two cases, we have chosen to make the distinction in order to add clarity to the proof.

Case I . n is even. Notice that, by (l), G has an odd number of loops in this case. Also note that, since the loop-coloring is admissible, at most 3 n distinct colors can occur on loops of G, and so the total number of distinct colors occurring in the edge-coloring of G is at most 1; + ~ ( n + 1 - ;)I = 1 7 1 5 n - 2 for n > 6. So only if n = 6 can more than n - 2 colors occur in G .

In the main part of the proof, the induction hypothesis will be invoked to complete

I

1 3n+2

ANDERSEN AND HILTON 222

a modification of G to Kfr-2. But before this part of the proof, we consider the case where the loop-coloring of G cannot be thus completed. We say that the loop-coloring of G is ( n - 2)-adrnissibk, if it is an admissible loop-coloring in a KA-2. Our first task then is to prove:

The edge-coloring can be completed if the loop-coloring of G is not (n - 2)-admissible. (*)

Proof of (*). Assume that the loop-coloring is not (n - 2)-admissible. Then the number t of colors occurring on an odd number of loops is equal to the number of uncolored loops. It is possible that G has n - 1 loops, in which case t = 1. Clearly, G must have at least loops. We first prove that we can extend the edge-coloring to all the uncolored loops of Kk so as to obtain an admissible loop-coloring.

Let cl, . . . , cI be the colors which need to be placed on one further loop, and let v1,. . . , vt be the vertices without prescribed loops. Let Ai be the set of colors from {CI, . . . , cr} which do not occur on any prescribed edge incident with v i (1 5 i 5 t ) . We wish to find a system of distinct representatives of (A1,. . . ,Af). Assume that this is not possible. Then, by P. Hall’s theorem [13] there is a set of k of the A; whose union has less than k elements, for some k , 1 5 k 5 t . Hence there must be at least t - (k - 1) colors each occurring at each of the k corresponding vertices of G. This uses at least k(t - k + 1) edges. G has n - t loops, and so it has 5 ( (a + 1) - (a - r)) =

3 (t + 1) edges. Therefore k ( t - k + 1) 5 t + 1, which can be rewritten

1

1

( k - 1) ( t - k ) 5 1 .

As 1 5 k 5 t , this implies that either k = 1, or k = t , or 2 = k = t - 1. If k = 1, then some A, is empty, so that each of the colors c1, . . . , ct is prescribed on an edge incident with vi, and so t 5 2 ( t + 1) implying t = 1; but then the only vertex without a loop is incident with an edge, contradicting (2). If 2 = k = t - 1 then t = 3, and two of the colors c1, c2, c3 are prescribed on edges incident with two vertices, contradicting that the number of edges of G is 2 in this case. Finally, if k = t , then some color occurs on edges incident with all t vertices without loops, again contradicting (2). It follows that the loop-coloring is possible, and we shall assume from now that it has been completed.

Let I’ be the number of free loops after completion of the loop-coloring. Recall that GI is the graph obtained from G by deleting all free Loops, their vertices and all isolated vertices. Then Gl is contained in a Kf with r = n - 1‘. Since the number of edges is at most 5 ( n + 1 - $), we also have r 5 21E(G)I 5 5 + 1. Let Z(i) be the number of free loops of color ci, 1 I i I R , after the completion of the loop-coloring, and let us note that if r = 2, then we can complete the edge-coloring to KL by Theorem 13, because 2r - n + E(i) = 4 - n + Z(i) which is positive only if E(i) = n - 3 or l ( i ) = n - 2; in the first case, ci occurs on a loop of Ki by the parity condition of an admissible loop-coloring, and in the latter case ci must be on both loops, or on the edge of K i , because otherwise G would have been of Type e3. So henceforth we shall assume that r > 2. We want to apply Corollary 17. With this in mind we do two preliminary steps in the particular case where r = 5 + 1 (then G has exactly 5 loops, all of distinct colors, r is even and at least 4, and the edges of G are independent; moreover, in the extension of the loop-coloring of G, each color occurs on exactly two loops); the first step is to obtain a loop-coloring with the property that at most one of the loop colors occurs on two free loops-this is easily achieved, because if two loop colors do, then there is a

1

1


color occurring on two nonfree loops, and one of these loops is not in G, and its color can be interchanged with one of the colors on two free loops; the second step is to take a 1-factor of K,, disjoint from E(G) , and give all its edges some color not used on G.

Now first assume that not all free loops have the Same color, let r n l and m2 be the numbers of free loops of the two colors occurring on most free loops, and let rn1 2 rn2; then at least ml - 1 and rn2 - 1 of these loops were loops of G, and putting rn = ml + m2 we see that G must have had at least I = [ ( rn - 2) + 5 (n - rn)] loops. Then (E(G)( is at most 1 (n + I - Z)], and letting y 2 0 denote 21E(G)I - r , a little calculation shows that this implies that 2r - n 5 -m + 6 - 2y, with equality only if n - rn = 2 mod 4. Notice that y = 0 implies that r is even and the edges of G are independent. To apply Corollary 17, we must make sure that each ci occurs with degree sum at least 2r - n + I ( i ) in Kf. From above it follows that

1

2r - n + I(i) 5 6 - 2y - (ml - Z(i)) - m2 5 5 . (9 If 2r - n + I(i) = 1 then E(i) is odd, so ci already occurs on a loop of Kf. If 1112 2 5 then 6 - 2y - (ml - I ( i ) ) - m2 5 6 - 0 - 0 - 5 = 1, so there is nothing to prove. So we need only consider the possibility that

2r - n + Z(i) 2 2 and m2 5 4 ($$I i) 1122 = 4. Then from ($1 and ($$)

so y = 0, rnl = Z(i) and 2r - n + I(i) = 2. Since rnl 2 m2 = 4 it follows that l ( i ) 2 4 and that r 5 5 (n - 2). It also follows that we have to place each such color ci on an edge of Kt if it is not already on an edge or loop of KE (since l ( i ) is even, if ci is on one loop it is on two). Since y = 0, IE(G)I = i. There are IE(KF)I - IE(G)I =

2 r(r - 1) - $ = 5 r2 - r edges available. There is no problem if there is only one such color ci. If there is more than one such ci then l ( i ) = ml = m2 = 4 for all such i. At most 7 colors can occur 4 times on the n - r free loops, and so up to 7 edges of Kf. - E(G) need to be colored. But since r = 5 ( n - 2) in this case and r 2 4 it follows that 5 r2 - r 2 - = "-'

1

1

1

I r+2 4 4 '

ii) m2 = 3. Then from ($) and ($$)

2 I 2r - n + l(i) 5 3 - 2 y - ( m l - I ( i ) ) 5 3 . ($$$)

Then y = 0, and ml = I(i) or rnl = l(i) + 1. Suppose first that Z(j) = ml - 1 for some color c j . Then m2 1 I ( j ) so either ml = m2 = 3 or ml = m2 + 1 = 4. If rnl = 4 then m is odd and so, by an earlier remark, we do not have equality in the middle inequality of ($$$), and so 2r - n + Z(i) 3 2 only for the one color c, with E(i) = r n l , and this color must occur in KF with degree sum 2, which is easily achieved. If ml = m2 = 3, we need to color an edge of Kf - E(G) for each color occurring on 2 or 3 free loops not already on an edge of G or on two loops of Kf; if the color occurs on 3 free loops, the edge must be nonincident with the vertex of Kf with a loop of the same color. At most 5 ( n - r - 2) edges of Kf - E(G) need to be colored. Since Z(i) is 2 or 3, it follows from ($$$) that 2r - n = 0, and so at most ( r - 2) edges need to be colored. Since Y L 4, 3 ( r - 1) ( r - 2) - 5 (r - 2) = 5 ( r - 2)2 2 f , and

1

1 1 1

ANDERSEN AND HILTON 224

so these edges can always be found, avoiding some vertex if necessary. Secondly suppose that no color cj has l ( j ) = ml - 1. If only one color ci occurs on ml free loops, then since 2r - n + Z( i ) 5 3 we only need one edge for this particular color. If more than one color occurs on in1 free loops, then ml = 1112 and the proof is as above.

iii) m2 = 2. Then from ($) and ($$),

2 5 2 r - n + I ( i ) S 4 - 2y - (ml - i ( i )) 54,

and so y = 0 or y = 1. 1 Assume first that y = 1. Then 2(E(G)( - r = y = 1, SO IE(C)l = T(r + 1). Also

so l(i) = ml and 2r - n + Z(i) = 2, so Z(i) is even and any such color c, needs to be put on an edge of Kf if it does not occur on an edge of G or a loop of K i . If there is only one such color this can be done, since IE(K,)( - (E(G)I = r ( r - 1) - 3 ( r + 1) = 3 (r2 - 2r - 1) 2 1 as r 2 3. If there is more than one such color then l ( i ) = ml = m2 = 2 and, from ($$$$), 2r - n = 0, and so we can assume that the number of such colors is at most L ? (n - r)] = Lij = 5 (r - 1). There are enough edges available since 3 r ( r - 1) - 3 (r + 1) 2 2 ( r - 1) as r 2 3.

1 1

1 1

1 I 1

Next assume that y = 0. Then

2 5 2r - n + L(i) 5 4 - (ml - Z(i)),

and r is even and the edges of G are independent. Suppose first that ml = 1122 = 2; then 2 5 2r - n + l(i) 5 2 + Z(i), so 2r - n is 0 or 2. But 2r - n = 2 would imply r = 5 + 1, and by our construction of the loop-coloring when r = 5 + 1 , ml zz 2 and m2 = 1, acontradiction.Therefore2r - n = 0, Z(i) = 2andhence2r - n + l(i) = 2, meaning only that such colors must occur on an edge of Kt if they do not already occur on an edge of G or a loop of K i . There are enough edges available for this, as there are at most 5 (n - r ) = such colors, and 3 r ( r - 1) - 5 > 2 as r 2 4. So suppose that rnl > m2. Since ml 2 3,2 5 2r - n + l(i) 5 1 + Z(i), so 2r - n 5 1, and since n is even we actually have 2r - n 5 0, so r 5 5. Now first suppose that ml = 3. If r < 5 then 2r - n + Z(i) 5 1 and nothing needs to be done. If r = :, the color on three free loops is also on an odd number of loops of Kf.; if it is on three or more loops of Kf. it does not need to be placed on an edge of Kf - E(G), but if it occurs on just one loop of KE (and no edge of G ) it needs to be placed on an edge of Kf not incident with the vertex of the loop, and this is possible because r 2 4. Up to a further 2 (n - r - 4) colors need each to be placed in an edge of Kf if they do not already occur in the K j , and this is possible since 5 r ( r - 1) - i - 1 > ( r - 4) = 5 (n - r - 4) as r 2 4. Secondly suppose that rnl = 4. Since n is even, if r < 5 we have 2 5 2r - n + I ( i ) 5 l(i) - 2, and the color on the 4 free loops may need to be placed on one edge, but no other color needs to be placed. This is easily achieved. If r = 5 then 2 5 2r - n + Z(i) = Z(i). The color of 4 free loops may require two edges of Ki - E(G) , or only one edge if it occurs on an edge of G or on two loops of KE. Since we are assuming that not all free loops have the same color, n - r > 4, so r 2 6. It follows that the two edges, or the one edge avoiding a given edge or two given loops, can be found in Kf. - E(G). Up to - 2 edges may be needed for further colors on

1 1 r r

1

1 1 1

1 (n - r - 4) =


1 two free loops. But 7 r (r - 1) - - 2. Finally suppose that mi 2 5. Then 2 5 2r - n + l(i) 5 l ( i ) - 1 and 2 5 2r - n + l(i) 5 4. Therefore only one color (the one with l(i) = m l ) may need to be placed on edges of Kt - E(G), and that on at most two edges. There is no problem if 2r - n + l(i) 5 3 so that we need at most one edge. If 2r - n + I(i) = 4 then, since not all free loops have the same color, r = (4 + n - (n - r - 1)) = 5 ( r + 5 ) , so r 2 5 , and since r is even, we get r 3 6. Therefore, as just above, we can find one or two suitable edges, as needed.

- 2 >

1 1 1 (4 + n - Z ( j ) ) L

iv) mz = 1. Since r 5 5 + 1 we have 2r - n + I(i) 5 2 + l(i). Assume first that r n l = m2 = 1. Then 2r - n + l(i) 5 3. If r # : + 1 then 2r -

n + E(i) 5 1 for all colors and there is nothing needing to be done. If r = + 1 then 2 5 2r - n + l(i) I 3 for each color c , and so each color which does not already occur with degree sum at least two needs to be placed on an edge of Kf - E(G), and each such edge may have to be nonincident with a vertex having a loop of the same color. The number of edges needed is at most n - 3, because in this case one color already occurs on the extra 1-factor added to Kf, one other color already occurs on two loops of K:, and at least one other color occurs on an edge of G. If r f 4, then r 2 6 and there are enough edges available avoiding a given vertex, because if r 2 8 then (r - 1) ( r - 2) - ( 5 - 1) - (i - 1) > n - 3, and if r = 6 the edges can be chosen first for those colors having vertices to avoid. If r = 4, we disregard the added 1-factor and consider the various cases, shown in Figure 10, directly, recalling that no color can occur on more than two loops in this case.

Next suppose that rnl > 1112. If r # 5 + 1 then 2r - n + l(i) 5 1 except for the one color on ml free loops, and so, except for this color, nothing needs to be done. For this one color we have, from ($) and ($$), 2 5 2r - n + l(i) 5 5 - 2 y , so y = 0 or y = 1. If y = 1 and Z(i) = ml then 2r - n + Z(i) I 3 and we may need to place c, on one edge of Kf. - E(G) if it occurs with degree sum 0 or 1, and we may have to avoid a loop. There is no problem except when r = 3 and 2r - n + ml = 3. But this case does not arise, for under these conditions mi = n - 3, so c, would occur on all the free loops, a contradiction. If y = 0 then r is even and r 2 4. We have 2r - n + l ( i ) 5 5. We may have to place c, on up to two edges, possibly avoiding the vertex of a loop, or on one edge, possibly avoiding up to three vertices. There is no problem if r 2 6, or if r = 4 and 2r - n + l(i) 5 3. The case r = 4 and 2r - n + Z(i) 2 4 does not arise, for under these conditions E(i) 2 n - 4, so all free loops have color c I , a contradiction.

Next suppose that m1 > rn2 and r = 5 + 1. Then r is even, r 2 4, y = 0 and each color occurs on exactly two loops in the extension of the loop-coloring. Therefore m1 =

2. We also have 2r - n + E(i) = 2 + l(i). Therefore we need to color up to two edges of K! - E(G) with one color, and possibly one edge for each of the remaining colors, and in some cases the edge must be chosen so that it misses some vertex. At least two of the desired edges are in the graph already, as one color occurs on the added 1-factor, and the other occurs on an edge of G. The required number of edges is therefore at most n - 1 . These are available, as T (3 + 1) 3 - ( $ + 1) 2 n - 1 for n 2 10, which is the smallest possible value for n in this case. Securing that 2 independent edges are chosen for one color, if necessary, and avoiding the loops for some other colors, is easy.

This completes the case where at least two colors are on free loops. Now assume that all free loops have the same color. Since G has an isolated

vertex, exactly n - r - 1 of the free loops were prescribed in G , so G had at least

I


Fig. 10. The Case R = 6, r = 4, rnl = m

8 8

2 1

8 8

2 1

2 1

Q Q 2 1

8 8 2 = 1

1 n - r - 1 + loops, and hence at most ~ ( n + 1 - ( n - r - 1 + [ $ I ) ) =

( r + 2 - 131) edges. Consequently r + 2 - 131 z r , so r 5 4. The case r = 3 cannot occur, because then G would have at least two edges and so at most n - 3 loops, and these would all be of the same color, as the color of the free loops must have

1

EVANS CONIECTURE 227

occurred on an odd number of loops of G; but then the loop-coloring would be (n - 2)- admissible, a contradiction. Since r > 2, it follows that r = 4. Therefore G has 2 edges, and n - 3 loops. Since at least n - (n - 3) = 3 colors occur on loops of G, it follows that G has n - 5 loops of the color of the free loops, and 2 further loops, one of each of two further colors. We need to extend the coloring of the edges of G and the loops of K i to an edge-coloring of Ki such that the degree sum in the Ki of the free loop color is 2r - n + (n - 4) = 4; also if ci is some other color then 2r - n + Z(i) = 8 - n, so there is no other requirement if n I 8; then the requirement can be met unless the free loop color already occurs on one edge of G , while the other edge has another color (all loops of K i have one of two other colors). This would imply, however, that G was one of the graphs of Figure 11, which can be dealt with as shown there. If n = 6 we get the same situations if the color of the free loops occurs on an edge of G, and we also have to make each color occur with degree sum at least 2 in Kf-but it is easy to extend the colorings of Figure 11 to this, coloring at the same time all edges of Kf. If the loop color

n=6: 3 i

2

Type e3

+ 3x2 i i , 3 1

Fig. 11. Revised Loop-Colorings for Certain Cascs.


does not occur on an edge of G, it can be added on two edges, and again each other color can be made to occur, unless G was as in Figure 11, last picture, dealt with as shown.

0 This completes the proof of (A).

By (*), we can now assume:

The loop-coloring of G is (n - 2)-adrnissible. (13)

Next we prove the following:

It suffices to prove that the edge-coloring can be completed if each edge of G has a loop on at least one of its end-vertices. (**>

Proof of (**). Assume that the theorem has been proved for graphs G with each edge having a loop on at least one of its end-vertices. We then prove it for any graph G:

Suppose that G has an edge e with no loop on either end-vertex. Let the color of e be c. By (13) we can suppose that the loop-coloring is (n - 2)-admissible. Let G’ be obtained from G by replacing e by a loop on each end-vertex, both having color c. Then the loop-coloring of G’ is admissible, and G’ cannot be a bad case, because if it were, then G would also be a bad case. We now observe that the edge-coloring of G can be completed if the edge-coloring of G’ can. For from a completion of the edge-coloring of G’, say with e receiving the color c’, we obtain a completion of the edge-coloring of G by giving e the color c and its loops the color c’. So we can proceed as follows. If the loop-coloring of G’ is not (n - 2)-admissible, we complete it by (*); otherwise, we use the same modification on G’ if it has an edge without loops on either end-vertex. Continuing in this way, we either obtain a completion of G, or we obtain a graph G’ which has a loop at one end-vertex of each edge, has (n - 2)-admissible loop-coloring, is not a bad case, and such that G can be completed if G’ can. This proves (**). 0

So we assume

Each edge of G has a loop on at least one end-vertex. (14)

Now let the number of free loops of G be If, and let r be the number of vertices of G I . By Lemma 21, G has at most ~ ( n + 1 - l f ) + lf nonisolated vertices. By (2), we know that this number is at most n - 1. We prove next:

2

The edge-coloring can be completed if G has exactly One isolated vertex. (***>

Proof of (***). If f (n + 1 - I f ) + If 2 n - 1 then If 2 n - 5. It follows that P I 4, and that if G has exactly one isolated vertex then G must be as in Figure 12.

In the last graph of Figure 12, the loop-coloring is not (n - 2)-admissible, so we only have to consider the two remaining cases. In both cases, since the loop-coloring is (n - 2)-admissible, there can be only one color occuring on an odd number of loops, and so this color can be placed on a loop on the isolated vertex, and a new color put on the remaining two loops. Then we can, in most cases, apply Corollary 17, with r = 4 for the first graph and P = 3 for the second.

TO deal with the second graph first, the condition of Corollary 17 is VaCUOUS for a C with 2r - n -k l ( C ) 5 0, i.e., l (c) 5 n - 6. The total number of free loops now

is n - 3, and if l ( C ) 2 n - 5 for Just one color c, then either l (c) = - 3 and c occurs On the original nonfree loop also (Since the loop-coloring was (n - 2)-admissible), in

EVANS CONjECTURE 229

Fig. 12. Graphs with Exactly One Isolated Vertex, n Even.

which case c can be put on the uncolored edge of the K3 as well, or Z(c) 5 n - 4 and again c can be put on the uncolored edge if it does not already color an edge of K3. In both cases, the assumptions of Corollary 17 are satisfied. If more than one color has Z(c) 2 n - 5 then 2 2, and so n = 6, implying, since the loop-coloring was (n - 2)-admissible, that one color c is on two free loops, another on one free loop and the original nonfree loop; putting c on an edge of the K3 if needed again makes the conditions of Corollary 17 satisfied.

For the first graph of Figure 12 we use the same approach. Assume first that n 2 8. The colors we need to attend to are those with Z(c) 2 n - 7. If just one color c occurs at least n - 7 times on free loops we argue as follows: if l ( c ) = n - 4, c is easily made to occur at all 4 vertices of the K4 except if it occurs on exactly one of the prescribed edges, in which case we interchange the color c on the unprescribed free loop and the color of the unprescribed loop on the edge not having color c; if l ( c ) = n - 5 then c also occurs on a prescribed loop in K4, and if it is not on an edge of K4 already it is placed on one; finally, if I(c) 5 n - 6 it can be put on an edge of K4 if needed. If exactly two colors occur on at least n - 7 free loops then either n = 10 and each occur on 3 free loops and one other loop, which leaves nothing to be done, or n = 8 in which case either one color is on one free loop and the other is on 3 and is easily given an occurrence on an edge of K4 if needed, or each occur on two free loops, needing degree sum 2 in K4 which is easily achieved. If at least three colors occur on at least n - 7 free loops then n = 8, one color is on two free loops, and two colors are each on one free loop. The color on two free loops is easily placed on an edge of K4. Finally, suppose that n = 6. Then we go back to G, with no extra loops colored, and obtain the K4 as

0 indicated in Figure 13. This completes the proof of (***).

Thus we can now assume

G has at least two isolated vertices. (15)

Before taking the main step of the proof for n even, we need a few more observations. One is that we want at most n - 2 colors occurring on G; we have already seen at the start of Case I that only if n = 6 can more than n - 2 colors occur, and from (15) it follows that this can happen only if G has 2 edges and 3 loops, all of distinct colors. But then the loop-coloring is not (n - 2)-admissible, contradicting (13). So we assume:

At most n - 2 distinct colors occur on G. (16)


'p 2 !21e 0

' p 2 8.l i 0

' p 3 f31i 0

' p 3 f i 0

'1 2 p1 i 0

'8 2 Y 3 1 i 0

'B 2 !12i 0

' p 3 Y12i 0

l f 2 f32e 0 ;p 3 p32i 0

4 'x2i ' 4 -+lX2i i +

4 5 4

5

Fig. 13. The Case Wherc G is the First Graph of Fig. 12 and n = 6.

It follows from (15) that r 2 3; it is also convenient to deal with the case r = 3 separately. If r = 3 we argue as follows. By (1) and (14) G has at least two nonfree loops. By (15) G has at most n - 5 free loops, and if it has exactly p1 - 5 free loops then G has exactly two edges and exactly two nonfree Loops. If G has exactly two nonfree loops then, since by (13) the loop-coloring of G is (n - 2)-admissible, and since by


(16) at most n - 2 colors are used, an unused color can be placed on the uncolored loop of K3, and the loop-coloring is still admissible. Since 2r - n + l(i) 5 0 for all i if G has fewer than n - 5 free loops, or if G has n - 5 free loops and they are not all of the same color, then in these cases we complete the edge-coloring of the K3, and then apply Theorem 13. If G has n - 5 free loops, all of the same color, then 2r - n + E(i) = 1 for one color c. Since there is an uncolored edge in K3 we can give this the color c if necessary; in any case, we complete the edge-coloring of K$ and then apply Theorem 13. Therefore we may suppose that

1

1

r 1 4 . (17)

If G has just one loop, say on the vertex v, then, by (I4), all edges are incident with v. It follows that they all have distinct colors, and completion is trivial. Since n is even, the number of loops is odd. So we assume that G has at least 3 loops, which implies:

(18) n - 2

2 G has at most - edges.

We now turn to the general argument. We divide this into two main cases; we first prove the following:

The edge-coloring can be completed if G has at most one free loop. (****) Proofof(****), In this case, we first assume that GI has no vertex of degree 1. Then n + 1 5 xvEV(Gl) d ( v ) 2 2r , so that r 5 5 . The number 2r - n + I ( i ) is nonpositive except for possibly one value of i when r = 4, for which it is possible that 2r - n + Z(i) = 1. In that casc we ensure that this one color occurs in Kn,2. In any case, we complete the edge-coloring of Ki keeping the loop-coloring admissible (there is always a color available for any uncolored edge or loop), and apply Theorem 13.

So in the following we assume that GI does indeed have a vertex of degree 1. We delete the edge e incident with a vertex of degree 1; let the color prescribed on E be 1. Removing two isolated vertices from G - e, we can consider i t as a subgraph of Kn-2; it still has an isolated vertex left, and so it can be a bad case only if it is of Type e3. If it is a base case of Type e3 then, by (I4), n = 6. Then, since r 1 4 by (17) and the loop-coloring is (n - 2)-admissible by (13), G is as in Figure 14, where a completion of K i is shown, which embeds, by Theorem 13, to a completion of G. Therefore in the following we assume that G - e is not a bad case.

So G - e is contained in a KA-2, has degree sum (n - 2) + 1 and is edge-colored with at most n - 2 colors; its loop-coloring is (n - 2)-admissible, and it is not a bad case. Then, by the induction hypothesis, we can complete the edge-coloring of G - e to an edge-coloring of KA-2. In this completion, let KE, be the complete graph, with loops,

1

I

3 1

! L o o -+3p 3 0 0 0

3

Fig. 14. The Situation when G-e is a Bad Case.


on the nonisolated vertices of G - e. Then r' = r - 1 plus the number of free loops (0 or 1) of G . By Lemma 21, r' 5 1: (n + l)] - 1 if G has no free loop, and r' 5 5 n] if G has a free loop. In both cases, an easy calculation shows that r' 5 7 [except if n = 6, r = 4 and G has a free loop, which is impossible by (IS)].

Let v be the end-vertex of e belonging to K:,; by (14) there is a loop of G on v, and clearly its prescribed color cannot be the color 1. If the color 1 occurs on an edge of Krl

incident with v, call this edge e'. The graph Kfr is embedded in an edge-colored KL-2, say with colors 1,2,. . . , n - 2. Therefore, by Theorem 13 with all l ( k ) equal to 0, each of the n - 2 colors occurs in K:I with degree sum at least 2r' - (n - 2) = 2r' - n + 2 in Krl. We want to modify the edge-coloring to make each of these n - 2 colors and an additional 2 colors each occur with degree sum at least 2r' - n . Since the loop- coloring of KEl is unaltered, and is (n - 2)-admissible, it is also n-admissible. Then, by Theorem 13, the modified edge-coloring of K:' can be completed to an edge-coloring of K i . When making the modification, we make sure that the colors of loops and edges of G are not changed, and that the color 1 does not occur at v after the modification; the completion will then be a completion of G. We obtain thc modified coloring by finding in K,, a path system with at least 2r' - n edges, all having distinct colors, and we recolor it with two new colors, say n - 1 and n, so that each occurs on at least 7 edges, i.e., with degree sum at least 2r' - n. As we lose degree sum at most two for each of the original colors, they also all occur with degree sum at least 2r' - n .

The existence of the path system is deduced from Corollary 20. We let F be the set of edges of G - e plus possibly one more edge to make it spanning: if G has a free loop, we add an edge joining v to the vertex of the free loop; otherwise, we add an edge incident with v if necessary. Then IF1 I IE(G)I 5 2, by (IS). We apply Corollary 20 (1) if 1 does not occur at v, and Corollary 20 (3) with M = {e'} if it does; in both cases

0

3n-4

2r'-n

n - 2

we get the required path system, and thus we have proved (****).

In the final argument we can therefore assume that G has at least 2 free loops. Then, by Lemma 21, r 5 17 (n - l)] . This is less than 7 except if n = 10, r = 6, and, by Lemma 21, G is as in Figure 15 [or n = 6 and r = 3 which is ruled out by (I7)]. But the graph of Figure 15 can easily be completed by Corollary 17: complete the loop- coloring so that each edge has loops of the same color on both end-vertices, and so that the four loops outside K: have two distinct colors, say 1 and 2, each on two of them; color a 1-factor of K: with a color not used on G , and finally make sure that 1 and 2 occurs with degree sum at least 4 in Kf, each other color with degree sum at least 2-there are plenty of edges available.

Hence we assume in the last step:

2

3n - 8 r 5 - 4 .

Fig. 15. A Graph Treated Separately.


We again want to find an edge e of G , such that G - e is not a bad case with respect to completion to KL-2 (by this we mean that, if the edge e is removed from G, and if two isolated vertices are also removed, then the resulting graph is not a bad case with respect to completion to Kt-2). We choose e at random; if G - e happens to be a bad case, it can have only one edge-otherwise G would contain an edge with no loop on an end-vertex; as r r 4, this edge is independent of e; since the loop-coloring is (n - 2)- admissible, deleting this edge instead of e gives a graph which is not a bad case, except if G is as in Figure 14; since we have shown how to complete this graph directly, we shall assume that e exists as required.

We then again have that G - e is contained in a KA-2, has degree sum (n - 2) + 1 and is edge-colored with at most n - 2 colors. Furthermore, G - e has a loop-coloring which is admissible in KL-2, and G - e is not bad case. Hence, by the induction hypothesis, we can complete the edge-coloring of G - e to an edge-coloring of KL-2 with n - 2 colors, say 1,. . . , n - 2. In this completion, naturally e gets some color, although not necessarily the correct prescribed color. Assume that this prescribed color of the edge e is the color 1. We shall now consider various cases depending on whether the color 1 occurs on edges or loops incident with the end-vertices of E or not. Clearly such occurrences of 1 are not prescribed. Since the color at at least one loop on an end-vertex of e is prescribed, 1 will not occur on loops at both end-vertices of e.

In all cases, we shall consider the edge-coloring of Kf C KL.2. We shall modify this edge-coloring of Kf so as to give e the correct color and make it possible to apply Theorem 13 to embed K, into K,, giving all the free loops the correct color. As usual for each color k, 1 5 k 5 n, let Z(k) be the number of free loops of color k. Then for the two new colors n - 1 and n, Z(n - 1) = Z(n) = 0. To apply Theorem 13, we must make all the colors occur with degree sum at least 2r - n + Z(k) in Kf. Let d(k) denote the degree sum of the color k in Kf, 1 5 k I n. Then, by Theorem 13, we have, with the present edge-coloring, in Kf:

d(k) 2 2r - (n - 2) + Z(k) = 2r - n + Z(k) + 2 ( 1 5 k 5 n - 2)

We also have

d(n - 1) = d(n) = 0.

The cases (a)-(f) that follow will also appear in the proof of the theorem for odd n. In the odd case, clearly the color 1 will not occur on the loop of both end-vertices of e; as explained above, this will not happen in the even case either.

CaseZ(a). The edge e gets the correct color 1. As F = E(G) is a spanning edge set of K, with at most edges [by (I8)], it follows from Corollary 20 (1) that K, has an (@,F)-d-system with at least 2r - n edges; this means that we can find a path system in K, containing at least 2r - n edges, all having distinct colors and not belonging to G. We recolor the edges of the system with the two colors n - 1 and n so that they both occur on at least edges. This reduces the degree sum of any other color by at most 2, so after the recoloring we have

d’(k) > 2 r - n + E(k) (1 I k I n),


where d’(k) denotes the degree sum of the color k in the new edge-coloring. We then apply Theorem 13 to obtain the required completion.

Case I(b). The edge e gets a color different from 1, say the color 2, and 1 does not occur at either end-vertex o f e . By Corollary 20 (l), with n - 1 in place of n, K , has an @,E(G))-d-system with at least 2r - n + 1 edges. Thus we can find a path system avoiding the edges of C , and with all edges having distinct colors, with at least 2r - n + 1 edges. If it contains an edge of color 2 we remove it, and so we have a path system with at least 2r - n edges and no edge of color 2. We recolor it with the two new colors n - 1 and n as before, and we give e the color I . Then we complete by Theorem 13.

Case I(c). The edge e is colored 2, and the color 1 occurs on edges el and e2, both incident with end-vertices ofe . We first temporarily recolor e with new color, n + 1, and we then give e2 the color 2. This may create a conflict, as the color 2 may already occur on an edge e’ incident with the end-vertex of c2 which is not an end-vertex of e (in applying Corollary 20, we disregard loops). If this is the case, we temporarily give e’ the color n + 1 . We let F be the spanning edge-set consisting of E(G) and e’ if it exists; then IF1 5 5. By Corollary 20 (4), K , contains an ({el,e*},F)-d-system with at least 2r - n edges, so we can find a path system including el and e2, and avoiding the edges of G and the edge el, with at least 2r - n edges all having distinct colors. We recolor it with the two colors n - 1 and n, each on at least edges, give e the color 1, and if e’ exists give it the color 2 again. Note that we do indeed obtain an edge-coloring with n colors, that e get the correct color, and that each old color, including 1 and 2, has its number of occurrences on edges reduced by at most 1. We finally apply Theorem 13.

Case I(d). The edge e is colored 2, and the color 1 occurs on an edge el incident with one end-vertex of e, and does not occur at the other. First we interchange the colors of the two edges e and el. If the color 2 already occurs o n an edge e’ incident with the end-vertex of el to which e i s not incident, we temporarily recolor e’ with a new color n + 1 . By Corollary 20 (3) , K , has an ({el},E(G) U {e’}]-d-system with at least 2r - n edges. We recolor as usual, let e keep the color 1 and give e’, if it exists, the color 2 again. Then we complete by Theorem 13.

Case I (e ) . The edge e is colored 2, and the color 1 occurs on a loop 11 incident with one end-vertex of e and does not occur ut the other. This is done as in Case I( b), except that after the recoloring we have to give I 1 a new color; any color not occuring at the vertex will do, as the loop-coloring we obtain by the induction hypothesis is ( n - 2)-admissible.

Case I( f). The edge e is colored 2, and the color 1 occurs on a loop 11 incident with one end-vertex of e and on an edge el incident with the other. By Corollary 20 ( 3 ) with n - 1 in place of n, K , has an ({el}, E(G))-d-system with at least 2r - n + 1 edges. If it contains an edge of color 2 we remove it, at so we have at least 2r - n edges to be recolored in the usual way with two new colors. We finally give e the color 1 and give 11

any available color. Then all degree sums are correct for an application of Theorem 13.

We have now completed the proof in Case I, even n.


Case IZ. n is odd. Throughout this proof, we let r denote the number of vertices of G,. Recall that we are assuming that n 2 7. The two main differences between the cases of even and odd n are that in the odd case there is less reason to worry about the admissibility of the loop-coloring, and that in the odd case 2r - n edges is not enough to make two colors occur with degree sum 2r - n each, since this number is odd. However we shall see that the fact that all loops have distinct colors makes it possible to include loops in a fruitful way in connection with our path systems.

We first prove:

The edge-coloring can be completed if G has no free edges and at most one free loop. (*****)

Proofof(*****). In this case it will be convenient to be able to distinguish between two cases by having E = 1 if G contains a free loop, E = 0 if not. Since G has no free edges, Gf denotes G with all isolated vertices deleted. By Corollary 23, we have that IV(Gf)l 5 7 + E ] 5 y. We begin by dealing with the situations where

First assume that n = 3 (mod 4). Then IV(Gf)l = y. We have that IV(Gl)l = 7 if G has no free loop, and IV(Gr)l = 7 if G has a free loop (and hence at least one other loop as the number of loops of G is even when n is odd and the degree sum of G is n + 1). By Corollary 23 this implies in the first case that all components of Gf are 2-paths, and in the second case that Gf has one free loop, one loop-edge pair, and the remaining components are all 2-paths. In both cases we then make a bipartition of G in the following way: Partition the vertices of Kf, into two sets, V(n+1)/2 having -y- vertices and V(n-1)/2 having vertices, in such a way that all edges of G join a vertex of V(a+lp to a vertex of V(n-lp, and such that each of the two loops, in the case where G has a free loop, goes in V(,,+1)/2 except if all edges and loops have distinct colors, in which case the nonfree loop goes in V(rL-lp. Thcre is no problem in doing this. Now add a new vertex v1 to V(n-1)/2, and replace each loop in Kf, by an edge to v1, keeping the color for the prescribed loops (See Fig. 16).

Now look at the subgraph of K(,+1) /2 , (n+1) /2 determined by G in this way. It has at most colors. We now apply

3(n + 1 - E ) ,= 3n+3

IV@f>l- 4 .

311-1 .

n + l

+ 1 edges, and they are colored with at most

Vn+l - 2

0 ov,

"E Vn+i 2 - 2

7 0 0 . . . . . ' 0 0 00::;

.___ If all colours distinct

To avoid bad case for n = 7

Fig. 16. Making a Bipartition of G.


Theorem 7 to complete this to an edge-coloring of K(n+1,n,(n+1)/2. Theorem 7 applies immediately if n > 9, for then we cannot have one of the exceptional situations, as our graph has maximum degree two and at most two free edges. The case n = 9 does not occur here since n = 3 (mod 4). If n = 7 and G is two 2-paths then the only bad case that could arise is Type 1; but this case can be avoided by putting both middle vertices in V(n-1)/2. If n = 7 and G has loops, we do not get a bad case if all colors are distinct; if they are not all distinct the graph formed might contain the bad case of Type 1, but this case can be avoided as just above unless the colors of the free loop and the edge of the loop-edge pair coincide, all other colors being distinct (then it also contains the bad case Type 2/3); in this case, we can choose the bipartition so that an edge of the 2-path belongs to V(n- l ) /~ , and we do not use the color of this edge. In every case we then color each of the K(n+1)~2’~ spanned by the vertex classes with the remaining colors. This is possible, because is even, and because if there is a precolored edge (from the nonfree loop or an edge of the 2-path) in one of the graphs, it has one of the remaining colors. Altogether we obtain an edge-coloring of Kn+l with IZ colors. We then remove the extra vertex, and replace the edges incident with it with loops. The edge-coloring of K,!, obtained in this way is clearly a completion of the edge-coloring of G.

Next assume that IV(G,)l = 7; then n = 1 (mod 41, n 2 9; recall that G has an even number of loops (and at most one free loop). We use a trick similar to the one above, but this time 7 is odd, which entails a slight change. Also, naturally the graphs are different. For clarity, we first present the argument for the case when G has no loops. Since IV(G,)l = (n + l), it follows from Corollary 23 that Gf has a component with more than two edges. If we could delete an edge e from such a component so that what was left was still connected, we would have degree sum n - 1, and then, by Corollary 23, the number of nonisolated vertices (which would not have changed) would be at most a (n - l), a contradition. So any such component is a tree. Deleting a vertex of degree 1 in such a component from Gf gives a graph with degree sum n - 1 and 7 vertices, which by Corollary 23 must consist of mutually disjoint 2-paths. It follows that G consists of isolated vertices, 2-paths, and one component which is either a 3-path or a 3-star. We make a bipartition of Kn as above making sure that all edges of G are contained in the I $ + I ) / ~ , ( ~ - . ~ ) D formed, and we add an extra vertex vl as above also. Then G is a subgraph of K(n+1)/2,(n+1),2 with edges, and if it is not a bad case with respect to completing to an edge-coloring of K(n+1)/2,(n+1)/2, we make such a completion by Theorem 7. The only way it could be such a bad case is when n = 9; then the bad case would be Type 1 and Gf would be a 3-star and a 2-path; but this is easily avoided by placing both centers in V(,-1)/2. Let the colors used be 1,. . . , 2; we may assume that at most one edge of G has color q. Now look at the two K(n+l)~2’~ spanned by the vertex classes. Color one of them with the q colors T,. . . , n, making sure that if the color occurs on an edge of G, then the color missing in the K(n+1)/2

at the end-vertex of this edge is q. Then color the other K(,+l)/z, this time making sure that the color missing at a vertex v always is the same as that missing at the other end-vertex of the edge of color q incident with v in the bipartite graph. Now the color occurs twice at many vertices. We obtain a proper edge-coloring of Kn+l by recoloring each edge of the bipartite graph of color q with the color missing at both its end-vertices (so that exactly one of these edges keeps the color 7, this being the one in G if there is one). We then finish as before, by replacing edges to vl by loops.

3n+l

<

3

3(n-1)

n + l


Now consider the case where G has exactly 2 loops. If it has no free loop, Corollary 23 implies that two components are loop-edge pairs, the rest are 2-paths. If G has a free loop, Corollary 23 again gives that one component of Gf is neither a free loop, a loop- cdge pair nor a 2-path, but that it has a vertex of degree 1, and if such a vertex is deleted then the components left are all of those standard types; it follows that Gf in addition to 2-path components and the free loop has either one 2-path with a loop on a vertex, or one loop-edge pair and one 3-path, or one loop-edge pair and one 3-star. In all cases, we can make the usual bipartition, introducing v~ and replacing each loop by an edge to vl; if all 7 + 1 edges and loops have distinct colors, we put a nonfree loop in the same class as v1, otherwise both loops go in the other class. This gives us a subgraph of K(,+1)/2.(~+1)/2 with at most 7 + 1 edges, colored with at most colors. Inspection shows that it can be a bad case with respect to completing to an edge-coloring of K(,+1)/2,(~+1)/2 only if n = 9, and that three of these cases (where G has two loop-edge pairs (Type 7/8), and where G has a free loop and a 2-path with a loop an an end-vertex (Type 7/8) or on the middle vertex (contains Type 1)) can be avoided by placing all vertices of degree 1 in V(n-1)/2 in the first case, and in V(n+1)/2 in the second case, and all vertices of degree 1 except the one with a free loop in V(,-1)/2 in the last case. There is one more situation where we get a bad case (containing a Type 1 and possibly Type 2/3), namely where G has a 3-star, for example with edges of colors 1,2, and 3, a loop-edge pair and a free loop; bad cases can occur if the loop-edge pair has colors (say) 5 and 4, and if the two loops have colors 5 and 4; but these can be avoided by changing the vertex classes of the 3-star, except if both situations occur simultaneously, so that the loop-edge pair have colors 5 and 4, and the free loop has color 4; in that case we let an edge of the 3-star be in V(n-lp. So we can assume that we do not have a bad case. We now finish as before; when coloring the two K(,+l)R’s, we first color the one with a precolored edge, if any.

This leaves us with the case where G has 4 loops, as by Corollary 23 more than 4 loops is not possible. Corollary 23 also gives that in this case Gf consists of one free loop, three loop-edge pairs, and 2-paths. Thus we have 2 + 2 edges and loops in this case. If they all have distinct colors, we put two loops in the same class of our bipartition as vl; the argument from above then goes through; bad cases do not occur. If not, and if some color occurs on exactly one edge or loop, we let both end-vertices of the corresponding edge be in the same class of the bipartition, with all other edges, coming from loops or not, being in the bipartite part. Again, the usual argument works (with no bad cases). Finally, if the colors that occur are on at least two edges or loops, we let the edge of the free loop be the edge not in K(n+1)/2,(n+1)/2; say this edge joins V I to v. Once more, we do not encounter bad cases. In this situation, we may assume that the color !$! is not used in G. We then complete the edge-coloring in the bipartite part in the usual way, and we do the same for the K(n+1)/2 containing the edge from the free loop, except that we ignore the color of this edge. Assume then that the edge receives the color n in the edge-coloring of K(,+l)n, and that the color it should have (from the prescribed color of the free loop) is 1. Let the edges in K(n+1)/2,(n+1)/2 of color 1 incident with v and v1 be vw and vlw’; neither of these cdges can be in G. Let the edges in K(n+1p(n+1)/2 of color 2 incident with w and w’ be wu and wu’ ({u, u’} may intersect {v,v,]). As 2 2 5 we may assume that the color n in the K(n+1)/2 now colored is not missing from any of u and u’. Therefore, when coloring the other K(n+1)/2, we can make sure that the color n is on the edge ww’, as well as having the missing colors correct. We then interchange the colors 1 and IZ in the 4-circuit vlvww‘, giving the correct color to the edge VIV. We finish by restoring the loops in the usual way.

n + l

n + l

n + l

n + l


3 n + l This completes the proof of (*****) for the case IV(G,)l 2 7, so we now assume that IV(Gf)I 5 y.

If G has no vertex of degree 1 (except possibly the one with the free loop), then n + 1 = d ( v ) 2 2lV(Gf)l - E , so IV(Gf)l 5 1 7 1 5 F. By Corollary 18, the edge-coloring can be completed: we only have to make each color occur on an edge or loop of K/I,(Gf)l if IV(Gf)l = q. For n > 7 there are enough uncolored edges and loops

for this, because when all loops have been colored we need at most edges which are not in G, and 7 7 (7 - 1) - 2 2 2 for n Z 9; if n = 7 the analogous inequality fails only if G has no loops (implying that G is a 4-circuit), and then we can choose the loop-coloring so as to have enough edges.

So we assume in the following that G has a vertex of degree 1, not incident with a free loop. If we delete such a vertex v1 and an isolated vertex [which exists by (2)], we obtain a graph with n - 2 vertices and with degree sum (n - 2) + 1. This graph G’ still has an isolated vertex, as IV(Gf)l I 7 so n - 2 - (IV(Cf)l - 1) 2 9 2 1, and it can be a bad case with respect to completion to Kn-2 only if n = 7, because G has no free edges. If n = 7 and G‘ is of Q p e 5b, then IV(G,)l = 4, and we can complete by Corollary 18; and if G’ is of Type 03, then it consists of two isolated vcrtices, a 2-path and two neighboring vertices both with loops, and we can select V I so as to avoid a bad case except if all edges and loops of G havc distinct colors; but then G has a trivial completion, because this configuration is contained in any edge-coloring of K7

with 7 colors. So in the following we assume that G’ is not a bad case, and we invoke the induction

hypothesis to complete it to an edge-coloring of Kn-2 with colors 1,. . . , n - 2. Let the edge incident with v1 be el, let the prescribed color of el be 1, let v be the other end- vertex of el, and let Kf, be the complete graph with loops spanned by all nonisolated vertices of G except q, that is, by the vertices of Gf - v1.

Since Kf, is embedded in an edge-coloring of KL-2 with n - 2 colors, by Theorem 13 each color k, 1 5 k 5 n - 2, occurs with degree sum at least 2r’ - n + 2 in KE,. We want to modify the edge-coloring, without changing the colors on loops and edges of G - v1, so that all colors, including colors n - 1 and n, occur with degree sum at least 2r‘ - n, which will enable us to complete to an edge-coloring of K i with n colors, again by Theorcm 13. We make sure that, in the modified coloring, the color 1 does not occur at v, so in the completion 1 will be on an edge joining I, to a vertex outside K , I , thus making it the desired completion of G.

When modifying the edge-coloring, we can recolor one edge of each color with a new color and still be sure of having the desired degree sum, 2r’ - n. Exactly how we carry out this recoloring varies according to the possible occurrence of the color 1 at v ; such an occurrence would have to be removed.

f l+l+&

I n + l n+l n+l n-1

Case (i). The color 1 does not occur at v. Let F be a spanning edge-set of K,, consisting of E(G)\{el} and, if this set is not spanning, an edge on v, joining v to the vertex of G with a free loop if G has a free loop, otherwise just joining v to any vertex. Note that the extra edge cannot be necessary if (E(G)( = q. Then IF1 5 2 - 1 = n--l 2 ‘ As r’ 5 y, it follows from Corollary 20 (l), with n - 1 in place of n, that K,, has an (@,F)-d-system with at least 2r’ - n + 1 edges, so we can find a path system in K,, avoiding all edges of G and with all edges having distinct colors. We recolor the edges

n + l


of the system with two new colors n - 1 and n such that both occur with degree sum at least 2r' - n + 1, which gives the required modification.

Case (ii). The color 1 occurs on a loop 8 on v. Let F be as in Case (i); we saw that (FI 5 2. and IF1 5 q, we can find in K,! a path system avoiding F and having at least 2 8 - n + 2 edges, all of distinct colors, by Corollary 20 (2) or (l), respectively, with n - 2 in place of n (except possibly in the first case if n = 7, but then r' = 3 and we can complete by Corollary 18). We now add the loop 8 to the system. We wish to obtain a path system and a loop on v, with v having degree 1 in the system, with all edges and loops having distinct colors, and with at least 2r' - n edges. To obtain this we may have to remove an edge of color 1 and an edge incident with v. Giving at least 7 of the edges left the color n - 1, and at least

If r' = 7 and IF1 = q, then G can have at most 2 loops, because IF1 5 IE(G)I. We consider the graph K,u obtained from K;, by adding a new vertex v" and replacing each loop of Kf. , by an edge to v" of the same color as the loop; as all loops have distinct colors, this gives an edge-coloring of K,u. We let F" consist of E(G)\{el} together with the edges corresponding to the loops of C if any, and if G has no loops also some edge incident with d', but not the one corresponding to 8. Then F" is spanning, and letting 1 denote the number of loops of G, we have IF'/] 5 7 - 1 + E + max{0,1 - 1) = n + l 3n-3 7. Furthermore, r" = __ . By Corollary 20 ( 3 ) with n + 1 in place of n, K,II has a ({vv"}, F")-&system with at least 2r" - n - 1 edges. As this number is 2r' - n + 1, this corresponds in K,! to a system consisting of at least 2r' - n - 1 edges and the loop e in addition to one more loop or edge. Notice that if there are two loops they are not joined by a path in this system. Therefore the system plus loops can be colored with two colors so that the loop-coloring is still admissible, and each new color gets degree sum at least 2r' - n, as required.

, we also construct K,II and F" as above. To estimate 1F"l in this case, we let 2 denote the number of loops of G and apply Corollary 23 on the size

n-1 3n-9 I 3 - 7 If r' I 7, or if r -

2r ' -n+ l

2r'-n - 1 edges and the loop the color n, we again obtain what we want. 3n-7

n t l - 1

3 - 5 Finally, if r -

of IV(G>I:

3n - 5 3(n + 1 - e ) - (1 - E ) + l - e < 4 4

n +5 implying 1 5 2e + 4 5 6. Therefore, 1F"l 5 2. In this case, r" = y. Since, by Corollary 20 (3) with n + 1 in place of n, K,// has a ({vv">,F")-d-system with at least 2r" - n - 1 edges, we can finish as before.

371-7 Case (iii). The color 1 occurs on an edge e incident with v. Tf r' 5 7, we let I; consist of the edges of ti except el, plus possibly two more edges needed to make F spanning: one incident with the vertex with the free loop, if it exists, and one incident with v if this was only incident with a loop and e l , in G-the edge joining these two vertices may be e, so we cannot rely on using it in F of course we only use two extra edges if E = 1. We get that IF1 5 q. By Corollary 20 (3) with n - 1 in place of n, in K,! we can find a path system with at least 2r' - n + 1 edges containing e and finish as usual. If 8 = 7 we finish as in Case (ii), except that the mandatory edge again is e.

0

3n-5

This completes the proof of (*****)


We therefore assume:

G has at least one free edge or at least two free loops. (113)

Next we prove:

The edge-coloring can be completed if G has exactly one isolated vertex. (******)

Proofof(******). We first note that we can assume that at most n - 2 colors are used on G; to see this,

note that the only graph of Figure 17 which contains n loops and edges is (e) if it has

In this case, G is the form of one of the graphs of Figure 17.

(h) - ]...I 8 . . . 8 0

Fig. 17. Graphs with Exactly One Isolated Vertex, n Odd.


no free edges; but in that case the edge-coloring has a trivial completion (or Theorem 13 can be applied with r = 2). The graphs of Figure 17 that have n - 1 elements are (b), (d), and (h) without free edges, and (e) with one free edge. Since n 2 7 and each color occurs on at most one free loop, it is easy to apply Corollary 17 to complete these graphs. So we assume that at most n - 2 colors are used on G.

We let ra denote the number of vertices of G I , rf the number of vertices of G j , and distinguish between three cases:

Case ( i ) . rl 5 y. There are at least 2 free loops, because n - 1 - - 4 = - 4 3

2. We delete two free loops, say of colors 1 and 2, to obtain a graph G’ C Kfi-2 with degree sum (n - 2) + 1. If G’ is a bad case with respect to completion to KL-2, it must be of Type 03, because it has exactly one isolated vertex, but then n - 3 must be at most 7, implying that n = 7. Then G is as in Figure 18, which is easily completed by Theorem 13 with r = 4 and the extended coloring shown.

So we can assume that G‘ is not a bad case, and by the induction hypothesis we can embed it in an edge-coloring of KA-2 with n - 2 colors, say 1,2,. . . , n - 2. We look at the KE, spanned by the vertices of G I . Each of the colors 1 and 2 occurs on a loop in KE, or on a loop on the isolated vertex; we may suppose that 1 does not occur on a loop on the isolated vertex, but that 2 may. We want to modify the edge-coloring of K:,, so that neither appears on a loop, and so that i t can be extended to an edge-coloring of Kf, with the colors 1,2,. . . , n. We do not change the colors on G , so the result will be a completion of G.

Let E(k) be the number of free loops of color k in G. Then, by Theorem 13, the color k occurs in Kf, with degree sum at least

3n-5 n+l

2rl - n + 2 + E(k) - 1 = 2r1 - n + 2

2rl - n + 2 + Z(k)

0 if k E {n - l , n } .

if k E {1,2},

and if k E {3,4 ,..., n - 2},

If the color 2 does not occur on a loop of Kf, , it occurs with degree sum at least 2r1 - n + 2 + E(2) = 2rl - n + 3. We want each color k to occur with degree sum at least 2rl - n + Z(k), and we see that we can delete the occurrence on an edge o f each of the colors 1,. . . , n - 2 except 1 and possibly 2 (if it is on a loop of Kf,), whose loop occurrences are all that we can delete. We now consider the complete graph on I-’ = rl + 1 vertices obtained from Kf, by adding a new vertex v1 and replacing each loop by an edge to vl of the same color as the loop; clearly we obtain an edge-coloring of K,I . We let F be the set of edges of K,, consisting of the edges of G1, and the edges incident with V I corresponding to loops of Gl; if G1 has no loops, we add to F an arbitrary edge incident with vl of color different from 1 and 2, to make F spanning. Since GI

3 2 2 i 5 i, 33 1 8 8 0 + 4

4 7

Fig. 18, The Graph C when G’ is a Bad Case (the Colors on the Free Loops of G are Not Assumed to be 1 and 2, as in the Text).


n + l 3n-1 3(n+l)-4 has at most two loops, IF1 5 2. As r’ = rl + 1 5 7 = ~ , by Corollary 20 (3) or (4) with n + 1 in place of n, K,! has a path system P avoiding all edges of F, containing the edge to vl of color 1 and that of color 2 if it exists, having all edges of distinct colors, and having at least 2r’ - (n + 1) = 2rl - n + 1 edges.

If P contains two edges incident with vl (which will be the case if both 1 and 2 occur on loops in K:,), we recolor the corresponding edges and two loops of K:L with the two colors n - 1 and n so that each gets on one loop and at least 7 (2rl - n - 1) edges; then each occurs with degree sum at least 2r1 - n and we can complete the coloring by Theorem 13. In any edge-coloring of KA, each color occurs on exactly one loop, and so in our case we do obtain the required completion of G. If P contains just one edge incident with vl, we recolor the corresponding loop of K:, with the color n - 1, give at least 1 1 T (2rl - n - 1) of the edges the color n - 1 also, and give at least T (2rl - n + 1)

edges the color n. Again, we complete by Theorem 13.

1

Case ( i i ) . t-1 5 F. > 2. We delete a free edge, say of color 1, to obtain a graph G”; if G” is a bad case with respect to completion to Kn-2 it must be of Type 0 3 , but since G is not a bad case not all free edges have the same color and so we can choose the deleted edge so as not to have a bad case, unless n = 7 and G is as in Figure 19. But this graph can be completed directly by Theorem 14 with r = 2, and so we shall assume that G” is not a bad case.

By the induction hypothesis, we can complete the edge-coloring of G” to an edge- coloring of Kfi-2 with colors 1,2, . . . , n - 2. Let the number of free edges in G of color k be e(k) , for 1 5 k I n (so that e (n - 1) = e(n) = 0). Then, by Theorem 14, each color k with 2 5 k 5 n - 2 occurs in K;, with degree sum at least 2rf - n + 2e(k) + 2, the color 1 occurs with degree sum at least 2rf - n + 2e(l), and colors n - 1 and n do not occur at all. Let F be a spanning edge-set of K , obtained by adding to E ( G f ) edges pairing those vertices having free loops in G, and possibly pairing one such vertex with any other vertex; as G has an even number of loops and at least 2 free edges, IF1 5 q. By Corollary 20 ( 1 ) with n - 2 in place of n, K , has an ( 0 , F ) d s y s t e m Q with at least 2rf - n + 2 edges. If Q contains an edge of color 1 we remove it from Q, and we color the remaining edges with the two colors n - 1 and n so that each gets degree sum at least 2rf - n + 1. Then we can complete the coloring by Theorem 14.

Then G has at least 2 free edges, because n - 1 -

3 n - 3 Case (iii). i-1 2 7 and rf 2 y. We let rg denote the number of vertices of G that are neither isolated, incident with a free loop nor incident with a free edge. It is seen from Figure 17 that rg is always 2, 3, 4, 5 or 6. As the number of vertices incident with free edges is even, we actually have that r[ 2 [TI + 8, where 6 is 1 if [TI - rg is odd, 0 otherwise. We then have

3 n - 3 3 n - 5 1 4 4 2

n - 1 = rl + r j - rg L ~ + ~ + - - rg + 8 ,

Fig. 19. The Graph G when G” is a Bad Case.

EVANS CONJECTURE

which implies

243

n 5 2(r, - 8) + 1.

As rg 5 6, this implies that n 5 11, and an examination of the various possibilities for rg shows that we have exactly the graphs of Figure 20 to deal with.

In view of (II3), G cannot be (4) or (6) of Figure 20. The remaining graphs can all be completed in the way used in the first part of the proof of (*****I, by making a bipartition of K, , adding a vertex v[ to take care of loops, and considering as many edges and loops of G as possible as contained in K ( , + 1 ) / 2 ~ ~ ~ + 1 ) / 2 . As in that proof, if there are !$! + 1 prescribed edges in the bipartite graph and all colors are distinct, we generally place a loop in the same class of the bipartition as vl. We only briefly comment on the situations where bad cases with respect to completing the edge-coloring of K ( n + l ) / ~ , ( n + 1 ) / 2 may occur:

8 Q

r = 3 ;

(141%- 1 Fig. 20. Graphs Satisfying the Incqualities of Case (iii).

8 8 o n = 9

o n = 7

0 n = 7

o n = 7

o n = 7

0 n = 7

8 o n = 7


The graph (3) could possibly give a bipartite graph of Type 10/11 of Figure 4; this is avoided by letting identically colored 2-paths have middle vertices in different classes. The graph (5) could be Type 2/3 with an edge added; in that case, a color on the 2-path is only used once, and we place the edge of that color in the class of vl. The graph (7) may be Type 1 with an edge added or Type 2/3 with an edge added; either possibility is avoided by placing an edge of the 3-star, alone of its color, in the class of vl. And (8) could be of Type 2/3 with an extra edge; but then one of the free loops has a color not used elsewhere, and we place it with vl. The graph (9) is the only case where there are + 2 edges and loops; if they all have distinct colors, we place the two nonfree loops in the class of v1, and if not then one of the loops has a color of its own and can go with vl-by choosing a free loop if possible, bad cases are avoided. The graph (10) can be of Type 1; it is avoided by placing an edge with vl. If the graph obtained from (12) is of Type 7/8, the free edge is alone of its color and is placed in the class not containing vl; if it is of Type 2 /3 with an edge added, a loop is alone of its color and is placed with vl. The graph (13) may be of Type 2/3 (and possibly Type 1 ) with an edge added; it is avoided by placing a nonfree edge, alone of its color, in the class not containing vl. Finally, (14) may be of Type 7/8 which is avoided by placing the free edge (alone of its color) with vl; and it may be of Type 2/3 with an extra edge, which is avoided either by placing the nonfree loop with vl or by placing the edge incident with the nonfree loop in the class not containing vl.

In all cases, we have avoided bad cases, and this completes the proof of (******). 0

We therefore assume:

G has at least 2 isolated vertices. (114)

Since n + 1 is even, the number of loops of G is even, and so it follows from (114) that G can have at most n - 3 loops. If it has n - 3 loops and 2 edges, all of distinct colors, completion is easy (Corollary 17 can be used). Therefore we can assume that G has at most n - 5 loops, and thus has at most n - 2 elements (loops and edges). So we can assume:

At most n - 2 distinct colors occur on G. (115)

Recall that r = lV(Gm)l. Let e and 1 be the number of free edges and free loops of G, respectively. The next step is to reduce the situation to graphs with r 5 7. So we first investigate graphs with r 2 y. Such a graph G can have at most one free edge and at most 2 free loops, because if we let x and y denote 2e and 1, with x 2 y , then, by Corollary 24, if x 2 3 we have

3n-5

3 n + 3 - 3 x + y 3 n + 3 - 2 x 3 n + 3 - 6

3n - 3

5 4

- x = I 3n + 3 + x + y

4 4 4

4 ’

r 5

- -

and one of the last two inequalities must be strict, because if x = 3 then y = 2e < x. So the largest number of 2e and I is at most 2.

Noting (I13), this means that we only have to consider the following possibilities for (e ,O: (0,2), (1,0), (1, I), and k 2 ) .


3(n- I) If ( e , 1) is (0,2) or (1,0), Corollary 23 gives that r 5 7, with equality only if each component of G that is not an isolated vertex, the free edge or one of the 2 free loops, is a 2-path. If (e , E ) is (1, l), Corollary 23 gives r 5 7 + 1 = y, with equality only if G has one loop-edge pair, the free loop and the free edge, and then only isolated vertices and 2-path components. If (e,Z) = (1,2), we get that r I 7 + 2 = 7, with equality only if all remaining components are 2-paths, but in this case it is also possible to have r = F: If Gl has loops, this is achieved only if it consists of one free edge, 2 loop-edge pairs and a number of 2-paths; if not, i t must contain a component with at least four vertices, and this component must contain a vertex of degree 1, whose deletion yields, by Corollary 23 again, a 2-path.

3(n-2)- 1

3 n - 3 ) 3n-1

It follows that if r 2 y, G is as one of the graphs of Figure 21. We note that due to (I14), we have n - r 2 4 for all the graphs of Figure 21. Since

r = 7 or r = 7 in each case, it follows that n 3 13. Then it is easy to see that they can all be completed by the bipartition method that we have used before; there are no bad cases with respect to completing the edge-coloring of K(n+1)/2,(n+1)/2, and we omit all details here.

3n-3 3n-1

For the final argument we can therefore assume:

3n - 5 t -5-

4 . It is clear that G , must have edges. Let e be an edge of G,n. Then, by (114) and (IIS),

removing two isolated vertices we can consider G - e to be a subgraph of K6-2 with

0 0 9 - 0

880-0 8 8 -

v -

0 * * * 0

Q-

0 0

880-0


degree sum (n - 2 ) + 1, colored with at most n - 2 colors. If possible, we choose e SO

that C - e is not a bad case with respect to completion to KA-2. If this is not possible, G is one of the graphs of Figure 22 [two further cases are ruled out by (113) and (IIS)]. The list of graphs of Figure 22 is easily obtained by considering any edge e’ of G,, assuming that G - e’ is a bad case, that is looking at Types 01 , 02 , 03, 5A, and 53 one by one, considering the various possibilities for e’ and seeing if the deletion of any other edge of G, gives a graph that is not a bad case.

All the graphs of Figure 22 are easily completed by Theorem 14 and, in one case, Theorem 13. So we assume in the following that G - e is not a bad case. We also make one further assumption: of the possibilities for the choice of e, we choose one whose color occurs on as few edges of G, as possible.

We then use the induction hypothesis to complete the edge-coloring of G - e to an edge-coloring of KL-2 with n - 2 colors. Just as for even n, we consider the complete graph with loops K: spanned by the edges and loops of G,. By Theorems 13 and 14, if C, = Gl each of the n - 2 colors occurs with degree sum at least 2r - n + 2 + E(k) in KL, and if G , = Gf with degree sum at least 2r - n + 2 + 2e(k), where Z(k) and e(k) are the number of free loops of color k and the number of free edges of color k, respectively. We modify the edge-coloring in the usual way, to make each of n colors occur with degree sum at least 2r - n + Z(k) or 27- - n + 2e(k), as the case may be.

Let the prescribed color of the edge e be the color 1 . We now have to consider the same cases as for n even, distinguishing between various possibilities for the occurrence of the color 1 at the end-vertices of e.

Case II(u). The edge e gets the correct color 1. We let F be a spanning edge-set of K , consisting of all edges of G in K, , plus additional edges joining disjoint pairs of vertices

3 2 o----o 0 0 n = 7 W l o r 3

1 2 o-----o 0 0

--?TO= 3 n = 7

Q4 Q 5 0 0 1 e 3

n = 7

n = 7

M 0 . 0 M 0 0 n r 7 1 2 1 1

Fig. 22. Thc Graph G when G - e is a Bad Case for any Choice of e.

247 EVANS CONJECTURE

of Kr with free loops in G if any, and an edge incident with the last such vertex if their number is odd. Then, by (II3), IF1 5 9. By Corollary 20 (l), with n - 1 in place of n, Kr has an (0, F)-d-system with at least 2r - n + 1 edges so we can find a path system in K , with at least 2r - n + 1 edges, all of distinct colors and not belonging to G. We recolor the edges of the system with the new colors n - 1 and n such that each occurs with degree sum at least 2r - n, and we then complete the edge-coloring to one of KL, using Theorem 13 or Theorem 14.

Case II(b). The edge e gets a color different fiom 1, say the color 2, and 1 does not occur at either end-vertex of e. In this case we need a number of diffcrent applications of our path system method, so we divide the proof into 3 subcases.

Suhcase 1. r 5 7; or r = 7 and either G, has degree sum less than n - 1 or it has at least 2 nonfree loops. Let F be a spanning edge-set of K, consisting of all edges of G,, plus a minimum set of additional edges to span vertices of G, with free loops. Then, in general, IF1 5 q, and if r = then the additional assumption implies IF1 5 7. By Corollary 20 (2) and (1) with n - 2 in place of n, K, has an (0, F)-d-system P with at least 2r - n + 2 edges (except possibly if n = 7, r = 3 and F is a K3; but then we can complete directly by Theorem 13 or Theorem 14). If some edge of P has color 2, we remove it from P, and we recolor the remaining system with colors n - 1 and n so that each occurs with degree sum at least 2r - n. We then give e the color 1, and we complete by Theorem 13 or 14.

and G, has degree sum n - 1 and at most one nonfree loop. In this case we consider the complete graph K,I obtained from Kf by adding a new vertex vi and replacing each loop by an edge to V I of the same color as the loop. As before, with all loops having distinct colors this gives an edge-colroing of K,.I. Let F' be a spanning edge-set consisting of all edges of G,, all edges corresponding to loops of G,, and, in the case when G, has no loops, one edge incident with v1. We claim that in K,., we can find a path system with edges of distinct colors avoiding F' with at least 2r' - n = 2r - n + 2 edges. Given this system, we can remove an edge of color 2 if it contains one, and be left with at least 2r - n + 1 edges in Kr, ; since there are at most two edges in the system incident with v ~ , the system now corresponds to edges and loops in Kf. with degree sum at least 2(2r - n), and since 2r - n + 1 is even, the edges and loops are colorable so that each of two new colors occurs with degree sum at least 2r - n. The required system exists, because since G, has degree sum n - 1 there is at most one free edge outside Kf, and so G, can have at most one free loop, implying if there is a free loop that K,I has at most 7 + 2 = forbidden edges (if G, has no loops, we made some edge incident with V I belong to F', but again we have only forbidden edges), and by Corollary 20 (2) it has an ((21, F')-d-system with at least 2r' - n edges (as r' = 7). As in the first subcase, we finish by giving e the color 1 and applying Theorem 13 or 14.

Then 3n = 4r + 5 = 1 (mod 4), so that n = 3 (mod 4). Note that n 2 11, because n = 7 would imply r = 4, contradicting (113) and (114). This is the most difficult situation to handle. Let If and I' be the total number of loops in G and in G,, respectively (also f and I are the number of free edges and free loops, respectively).

If 2f > 1 then G,n = Gf and all free loops of G are in G,, so I + = E'. By Corollary 24,

3n-9

n-3

Suhcase 2. r =

(n-1)-2

3n-5 Suhcase 3. r = 7.

24%

3n - 5 -= - 2 f , IV(GnI)l 5 4

3(n + 1) + 2f + 1 4

ANDERSEN AND HILTON

so 6 f - E 5 8. Therefore f = 1, and I = 0 or 1 = 1. Thus the number of free loops in G, is at most one, and it can only equal one if there is exactly one free edge outside G,.

If E 2 2f then (113) implies that 1 2 2. We have that G, = Gi and all free loops of G are outside G,, so 1' = E + - 1. By Corollary 24,

3n - 5 4

3(n + 1) + 2f + 1 4 - 1 , -- - IV(Gm)I 5

so 21 5 31 - 2f 5 8, implying that either f = 0 and 1 = 2, or f = 1 and 1 E {2,3}, or f = 2 and I = 4.

Thus taking these two cases together f E (0, 1,2}, if f = 1 then E E {0,1,2,3}, and i f f = 2 then I = 4. Now let G' be the graph G with all free edges and free loops, together with all incident vertices and isolated verties, removed. The number of loops in G' is 1' - 1.

If 2f > 1 then IV(G')l = IV(G,)l - E , so by Corollary 23,

3n - 5 4

3(n + 1 - 2f - 1) - ( E + - I ) ~- 4

1 = IV(G')l 5 1

so 1' = 1' I 21 + 8 - 6 f . Consequently we have the possibilities:

(A) f = 1, I = 0, 1' 5 2; (B) f = 1, I = 1, 1' E {2,4} (as 1' must be even here).

If 1 2 2 f then IV(G')l = IV(G,)l - 2f, so by Corollary 23,

3n - 5 3(n + 1 - 2 f - 1 ) - ( E + - I ) ___- 2f = IV(G')l 5 4 4 ,

so 1' 5 8 + 2f - 21 and hence 1' 5 8 + 2f - 31. Consequently we have the further possibilities:

(r) f = 0, 1 = 2, 11 5 2;

(Z) f = 2, 1 = 4, 1' = 0.

(A) f = 1, 1 = 2, I' I 4; ( E ) f = 1, 1 = 3, E' = 1 (as 1' must be odd here);

Notice that the number of loops within G, is at most 4. For the path system argument, it turns out that we need to know that some edge el of

K, of color 1 is not in G, the reason for this will become apparent further on. Suppose first that at least 2 distinct colors occur on edges of G,. Let c be a color occurring on fewest edges of G,; by the choice of e, the color 1 occurs on as few edges of G, as c, unless the deletion of any edge of a color with the least number of occurrences would give a bad case with respect to completion to Kfi-2. So suppose that this is the case, let e, be an edge of color c in G, and assume that G - e, is a bad case. Recall that in this subcase, r = 7.

First suppose that G - e, contains a Type 01 graph. Then G - e, consists of some free loops, some free edges (possibly none), one edge with an attached loop,

3n-5


and two isolated vertices. Clearly IV(G')l 5 6. We have IV(G')l + 1 + 2f + 2 = n. If IV(G')l = 6 then I is odd and we must have case (B) above-case (E) is ruled out by the parity condition on n; hence the only possibility is the graph 1 of Figure 23. If IV(G')l = 5 then I is even, and only case (A) is consistent with r = y; hence G must be the graph 2 of Figure 23. If IV(G')l = 4 then I is odd, but neither case (B) or case (E) is consistent with the value of Y. Likewise, there are no possibilities with IV(G')l = 3.

If G - e, contains a Type 0 2 graph, then G - e, consists of one free loop, one edge with a loop attached, some free edges and two isolated vertices. With one exception the edges are colored the same. As above, IV(G')l 5 6, and in this case, due to the single free loop, r 5 7. As n 2 11, these observations and r = 7 imply that n = 11, r = 7 and IV(G')l = 6. Then the graph G is as in 3 of Figure 23, and it must have the coloring shown, because if the exceptional edge is not free, it can be deleted without creating a bad case, contrary to hypothesis.

If G - e , is a Type 03 graph we immediately get r 5 6, which is impossible as n 2 11. Also, G - e , cannot be of Type 5a or 56.

The graphs of Figure 23 are easily completed by coloring the KL in 1 and 3, and the Ki in 2, spanned by the vertices of G', in such a way that a completion by Theorem 13 is possible and necessarily contains the required configuration.

Thus, still assuming that not all edges of G , have the same color, we can now assume that the color 1 of e occurs on at most half of the edges of G,, that is on at most edges.

- 1 edges of G, - e, and by Theorem 13 it appears on at least edges of K,; but 7 > 4 as r > 7- So let el be an edge of color 1 but not in G. Let K,I be obtained in the usual way by adding vl and replacing loops by edges to Q, and do the following: recolor el with the color 2. If there are edges incident with the end-vertices of el having color 2, temporarily recolor them with a new color n + 1. L A F consist of all edges of G,, of the edges corresponding to loops of G,, of the edges of color n + 1, and, if G, has no loops, of some edge incident with vl. Then IF1 5 7 + 4 + 2 = F. By Corollary 20 (3) with n + 1 for n, Krr has an ({e,},F)-d-system with at least 2r' - n - 1 edges, so we find a path system as usual, containing the edge el and avoiding all edges of F, having at least 2r - n + 1 edges. The path system will not contain any edge whose original color was 2, but it may contain two edges whose original color was 1. However, when we recolor the edges of the path system in the usual way, give e the color 1 and recolor edges of color n + 1 with their original color 2, we get what we require to embed by Theorem 13 or 14.

3n-5 .

Then the color 1 occurs on at most Zr-(n-2)-1 Zr-n+l n-5 3n-7

n-1-4

1.0

2. 0

3. 0

0 M n2 n = l l , r = 7 1 1 1 1

n = 11, r = 7 I

0 I

0 - c o M n = 1 1 , r = 7 eC I 1 2 1

0 ' 8 Fig. 23. Graphs G Where Delction of e , Gives a Bad Case.


To finish subcase 3, we must consider such edge-colored graphs G for which all edges of G, have the same color c’. If all edges of the whole graph G have the same color, thcn the edges are independent, so any edge-colored KA using colors 1,. . . , n embeds G. So we can assume that G has at least one free edge outside G,. But, as we saw in the beginning of this subcase, then G must have precisely one free edge, of color c” # c’, and it can therefore have at most one free loop. If it has no free loop, then by (A) I’ 5 2, and as the edges of G, are independent each must have a loop at an end-vertex, and so r 5 4, a contradiction. So G has exactly one free loop, and by (B) we get 2’ = 4, which implies n = 11. Then G can be completed in the following way: If the color of the free loop is c’, the color the K i spanned by the free edge and 2 isolated vertices so that c’ occurs on 2 edges, c” occurs on a loop if it is not on a loop of G and no color from a loop of G occurs on a loop; embed by Theorem 13. If the color of the free loop is not c’, then just color the K i spanned by the free loop and edge, such that no color on a nonfree loop of G goes on a loop; again, embed by Theorem 13.

This finally completes the proof of subcase 3 and so of this case.

CaseZZ(c). The edge e is colored 2 and the color 1 occurs on edges el and e2, both incident with end-vertices of e. We temporarily recolor the edge e2 with the color 2, and give the color n + 1 to e, and if there is an edge er of color 2 incident with the other end-vertex of e2 this is recolored n + 1 as well. We then consider the spanning edge-set of K,. consisting of the edges of G,, together with e’ if it exists, and a minimum set of additional edges to span the vertices with free loops in G,. Then IF1 5 2 + 1 = - 2 ‘

By Corollary 20 (4) with n - 1 for n, if r 5 then K , has an ({el, e2}, F)-d-system with at least 2r - n + 1 edges. Recoloring the system in the usual way, giving e the correct color I, and if needed giving e’ its original color 2, we obtain a situation where Theorem 13 or 14 can be invoked to give a completion of the edge-coloring of G.

If r = y, then by the argument in Case II(b), Subcase 3, G, has at most 4 loops. We then add a vertex vl to KE and let K,! be the complete graph on r + 1 vertices obtained by replacing loops by edges to vl, and we let F’ be the edges of G,, d if it exists, and the edges corresponding to loops of G,; if G, has no loops, we add an edge incident with vl. Then IF‘I 5 7 + 4 + 1 = F. By Corollary 20 (4) with n + 1 in place of n, K,! has an ({el , ez}, F’)-d-system with at least 2r’ - n - 1 edges, so there is a path system of the usual kind. containing el and e2, having at least 2r - n + 1 edges. We finish as above.

Case ZZ(d). The edge e is colored 2, and the color 1 occurs on an edge el incident with one end-vertex of e, and does not occur at the other. We interchange the colors 1 and 2 on el and e; if 2 occurs on an edge e’ incident with e l , we temporarily recolor e’ with a new color n + 1. If r 5 y, we find a path system with at least 2r - n + 1 edges, containing el (now having color 2) and avoiding the edges of G, and e‘ if it exists (and possibly some further edges incident with vertices of G, with free loops). We can do this, because the set F of forbidden edges has IF1 5 7 and so, by Corollary 20 (3) with n - 1 for n, K , contains an ({q},F)-d-system with at least 2r - n + 1 edges. If r = y, we add a vertex v[ in the usual way and obtain a forbidden set of edges of size at most F. By Corollary 20 (3) with n + 1 in place of n, the extended graph K,! has a path system containing el with at least 2r’ - n - 1 edges, and we finish as usual, letting e keep the color 1 it obtained after the interchange, and recoloring (possibly) d

n-1 n+l

n-1-4


with the color 2. The edge el, belonging to the path system gets one of the new colors n - 1 and n.

Case II(e). The edge e is colored 2, and the color 1 occurs on a loop 11 incident with one end-vertex of e and does not occur- at the other. If r 5 y, then, by Corollary 20 (3) with n - 1 for n, we can find a path system of 2r - n + 1 edges with all edges having distinct colors, containing e and avoiding all other edges of G,, as well as some edges added to obtain a spanning set of forbidden edges: up to 2 edges incident with the end- vertices of e and some edges incident with vertices of G, with free loops. Removing e from and adding ZI to the path system, we can still recolor it with two new colors n - 1 and n so that both get degree sum at least 2r - IE. We do this, and give e its correct color 1. Then we complete the edge-coloring by Theorem 13 or 14.

If r = y, we add a loop-vertex v~ in the customary way. The situation obtained is then exactly that of Case II(d), so the completion takes place much as in the case, with e recolored 1 and 11 recolored n - 1 or n.

Case I f ( f ). The edge e is colored 2, and the color 1 occurs on a loop 11 incident with one end-vertex of e and on an edge el incident with the other. In K,, interchange colors on e and el; the color 1 did not occur-in K,-on the vertex of 11, but if the color 2 occurred at the end-vertex of el not incident with e, we temporarily recolor the edge of color 2 with a new color n + 1. We want a path system of 2r - n + 1 edges containing both e and el; before recoloring, we replace e by 1 1 . To make a spanning set of forbidden edges, we may have to add two edges, incident with the end-vertices of e, as well as edges pairing off free loops. Naturally, the edges of G, different from e, and the edge of color n + 1, are forbidden. We get at most forbidden edges. By Corollary 20 (4) with n - 1 for n, if r 5 y, we can find the required system. We end up by recoloring e with the color 1, 11 and el with n - 1 or n, and the possible edge e’ is given its original color 2.

we add a vertex v f as before. The situation is then as in Case II(c), so we handle it as we did in that case (that is, by giving e and possibly some other edge the color n + 1, giving the edge e2 corresponding to 11 the color 2, and asking for a path system containing el and e2, and K,, contains a ({e1,ez},F)-&system of at least 2r’ - n - 1 edges, where F is the set of forbidden edges).

13

If r =

This completes the proof of Theorem 8.

REFERENCES

[l] L. D. Andersen, Embedding latin squares with prescribed diagonal, Ann. Discrete Math. 15

121 - , Completing partial latin squares, Mat.-Fys. Medd. Danske Vid. Selsk. 41:l (1985),

[3] -, Hamilton circuits with many colours in properly edge-coloured complete graphs, Math. Scand. 64 (1989), 5-14.

[4] L. D. Andersen, R. Hlggkvist, A. J. W. Hilton, and W. B. Paucher, Embedding incomplete lalin squares in latin squares whose diagonal is almost completely prescribed, European J . Combin.

(1982), 9-26.

23-69.

1 (1980), 5-7.


[5] L. D. Andersen and A. J. W . Hilton, Thank Evans!, Prac. London Math. SOC. (3) 47 (1983),

[6] -, The existence of symmetric latin squares with one prescribed symbol in each row and column, Ann. Discrete Math. 34 (1987), 1-26.

[7] -, “Extending edge-colorings of complete graphs and independent edges,” in Graph Theory and its applications: East and west, proceedings of the first China-USA international graph theoly conference (1989), 30-41.

[8] L. D. Andersen, A. J. W. Hilton, and C. A. Rodger, A solution to the embedding problem for partial idempotent lalin squares, J . London Math. Soc. (2) 26 (1982), 21-27.

PI - , Small ernbeddings of incomplete idempotent latin squares, Ann. Discrete Math. 17 (1983), IY-31.

[lo] A. B. Cruse, On embedding incomplete symmetric latin squares, J . Combin. Theory Ser. A

[ l l ] R. M. Darnerell, On Smetaniuk’s construction for lafin squares and the Andersen-Hilton

[12] T . Evans, Embedding incomplete latin rectangles, h e r . Math. Monthly 67 (1960), 958-961. [13] P. Hall, On representations of subsets, J . London Math. SOC. 10 (1935), 26-30. [14] A. J . W. Hilton, Embedding an incomplete diagonal latin square in a complete diagonal latin

~ 5 1 - , Embedding incomplete double diagonal latin squares, Discrete Math. 12 (1Y75),

[16] A. J. W. Hilton and C. A. Rodger, Latin squares with prescribed diagonals, Canad. J . Math.

[ 171 D. Hoffmann, Completing incomplete commutative lafin squares wcth prescribed diagonals,

[18] C. C. Linder, Embeddingpartial idempotent latin squares, J . Combin. Theory Ser. A 10 (1971),

[19] C. A. Rodger, “Embedding incomplete idempotent latin squares”, Combiizutorial mathematics X, L. R. A. Casse (editor), Proceedings of the conference held in Adelaide, Aus- tralia, August 23-27, 1982; Lecture notes in Mathcmatics 1036, Springer, Berlin, 1983,

[20] - , Embedding an incomplete latin square in a latin square with u prescribed diagonal,

[21] B. Smetaniuk, A new construction on latin squares-I: A proof of the Evans conjecture, Ars

507-522.

16 (1974), 18-22.

theorem, Proc. London Math. SOC. (3) 47 (1983), 523-526.

quare re, J . Combin. Theory Ser. A 15 (1973), 121-128.

257-268.

XXMV (1982), 1251-1254.

European J. Combin. 4 (1983), 33-35.

240- 245.

pp. 355-366.

Discrete Math. 51 (1984), 73-89.

Combin. XI (1981), 155-172.

Received March 9, 1993 Accepted June 24, 1993

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Symmetric latin square and complete graph analogues of the evans conjecture